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Karnataka Board Class 9 Maths Chapter 12 Circles Ex 12.2

Question 1.
Recall that two circles are congruent if they have the same radii. Prove that equal chords of congruent circles subtend equal angles at their centres.
Solution:

Data: Two circles having centres O and O’ and having equal radius. Chord AB = CD.
To Prove: Equal chords of congruent circles subtend equal angles at their centres.
Or ∠AOB = ∠CO’D
Proof: In ∆AOB and ∆CO’D,
OA = O’C (Data)
OB = O’D (Data)
AB = CD (Data)
∴ ∆AOB ≅ ∆CO’D
∴ ∠AOB = ∠CO’D.

Question 2.
Prove that if chords or congruent circles subtend equal angles at their centres, then the chords are equal.
Solution:

Circle with centres A and B are congruent and ∠PAQ = ∠RBS
To Prove PQ = RS.
Proof: In ∆APQ and ∆BRS,
AP = BR ……… radii of congruent circles
AQ = BS are equal
∠PAQ = ∠RBS (Data)
∴ ∆PQR ≅ ∆BRS (SAS postulate)
∴ PQ = RS.

KSEEB Solutions for Class 9 Maths

 

KSEEB Solutions for Class 9 Maths Chapter 8 Heron’s Formula Ex 8.2 are part of KSEEB Solutions for Class 9 Maths. Here we have given Karnataka Board Class 9 Maths Chapter 8 Heron’s Formula Exercise 8.2.

Karnataka Board Class 9 Maths Chapter 8 Heron’s Formula Ex 8.2

Question 1.
A park, in the shape of a quadrilateral ABCD, has ∠C = 90°, AB = 9 m, BC = 12 m, CD = 5 m and AD = 8 m. How much area does it occupy?
Solution:
BD is joined.
In Right angled ABCD, ∠C = 90°

As per Pythagoras Theorem,
BD² = BC² + CD²
= (12)² + (5)²
= 144 + 25
BD² = 169
∴ BD = 13m.


Now, in ∆ABD,
Let a = 9m, b = 8m, c = 13m.

∴ Area of quadrilateral ABCD,
= Area(∆ABD) + Area (∆BCD)
= 35.46 + 30
= 65.46 sq.m.
Area of quadrilateral ABCD = 65.46 sq.m.

Question 2.
Find the area of a quadrilateral ABCD which AB = 3 cm, BC = 4 cm, CD = 4 cm, DA = 5 cm, and AC = 5 cm.

Solution:
In quadrilateral ABCD, AC is the diagonal.
In ∆ABC, a = 5cm, b = 3 cm, c = 4 cm.
∴ Perimeter = a + b + c = 5 + 3 + 4 = 12 cm.


∴ Area(∆ABC) = 6 sq. cm.
In ∆ADC, a₁ = 5cm, b₁ = 5 cm, c₁ = 4cm.
∴ Perimeter = a₁ + b₁ + c₁
= 5 + 5 + 4 = 14 cm.

Area of ∆ADC = 9.16 sq.cm.
∴ Area of quadrilateral aBCD,
= Area(∆ABC) + Area(∆ADC)
= 6 + 9.16
= 15.16 sq. cm.

Question 3.
Radha made a picture of an aeroplane with coloured paper as shows in Fig. Find the total area of the paper used.

Solution:
(I) ∆ABC is an isosceles triangle.
AB = AC = 5 cm, and BC = 1 cm.
∴ a = 5 cm, b = 5 cm, c = 1 cm.
∴ Perimeter = a + b + c = 5 + 5 + 1 = 11 cm.

∴ Area(∆ABC) = 24.85 sq. cm.

(II) PBCS is a rectangle.
l = 6.5 cm, b = 1 cm.
Area of rectangle PBCS = l x× b
= 6.5 × 1 = 6.5 sq.cm.

(III) PQRS is a trapezium PT ⊥ QR
In Right angle ∆PTV,

(IV) Area of ADBE = Area of ∆FCG

(V)

∴ The total area of the paper used to make an aeroplane
= 2.485 + 6.5 + 1.29 + 4.5 + 4.5
= 19.275 sq. cm.

Question 4.
A triangle and a parallelogram have the same base and the same area. If the sides of the triangle are 26 cm, 28 cm and 30 cm, and the parallelogram stands on the base 28 cm, find the height of the parallelogram.

Solution:
Perimeter of ∆ABC = 26 + 28 + 30 = 84 cm.

Question 5.
A rhombus shaped field has green grass for 18 cows to graze. If each side of the rhombus is 30 m and its longer diagonal is 48 m, how much area of grass field will each cow be getting ?

Solution:
ABCD is a rhombus.
Area(∆ABD) = Area(∆BDC).
In ∆ABD, a = 30 m, b = 30 m, c = 48 m.
∴ Perimeter= a + b + c = 30 + 30 + 48 = 108 m

But, Area of whole field = 2 × area of ∆ABD
= 2 × 432
= 864 sq.m.
∴ Are required for growing grass for 18 cows is 864 sq.m.
∴ Area required for growing grass for 1 cow ………. ??
\(\frac{864}{18}\) = 48 sq.m.

Question 6.
An umbrella is made by stitching 10 triangular pieces of cloth of two different colours, each piece measuring 20 cm, 50 cm and 50 cm. How much cloth of each colour is required for the umbrella?

Solution:
Each sides of triangular shaped cloth
Let a = 50 cm, b = 50 cm, c = 20 cm.
∴ Perimeter = a + b + c
= 50 + 50 + 20 = 120 cm.


∴ Total Area of Umbrella mad up with 10 triangular shaped cloth is

Question 7.
A kite in the shape of a square with a diagonal 32 cm and an isosceles triangle of base 8 cm and sides 6 cm each is to be made of three different shades as shown in Fig. How much paper of each shade has been used in it ?

Solution:
Kite is in the shape of square.


∴ Total area = I + II + III
= 256 + 256 + 17.92
= 549.92 sq.cm.

Question 8.
A floral design of a floor is made up of 16 tiles which are triangular, the sides of the triangle being 9 cm, 28 cm and 35 cm. Find the cost of polishing the tiles at the rate of 50p per cm².

Solution:
In the given figure there are 16 triangles.
Each sides of triangle are
a = 28 cm, b = 9 cm, c = 35 cm.
Perimeter of triangle = a + b + c
= 28 + 9 + 35 = 72 cm.

∴ Area of 16 triangles = 16 × 88.2 = 1411.2 sq.cm.
Polish expenditure for 1 sq.cm. = 50 p = Re. 0.5.
∴ Polish expenditure for 1411.2sq. cm. =?
= 1411.2 × 0.5
= Rs 705.6

Question 9.
A field is in the shape of a trapezium whose parallel sides are 25 m and 10 m. The non-parallel sides are 14 m and 13 m. Find the area of the field.

Solution:
ABCD is a trapezium.
AB || DC.
Draw AE || BC and AF ⊥ DC.
∴ ABCE is a parallelogram.
DE = DC – EC = 25 – 10
DE = 15 cm.
Sides of ∆ADE
a = 14 cm, b = 13 cm. c = 15 cm.
∴ Perimeter= a + b + c
= 14 + 13 + 15 = 42 cm.
∴ s = \(\frac{a+b+c}{2}=\frac{42}{2}=21 \mathrm{cm}\)
∴ Area of ∆ADE, A

(ii) Area of quadrilateral ABCE = base × height
= EC × AF
= 10 × 11.2
= 112.0
= 112 sq.cm.
∴ Complete area of trapezium ABCD, A
A = Area of ∆ADE + Area of quadrilateral ABCE
= 84 + 112
= 196 sq.cm.

We hope the KSEEB Solutions for Class 9 Maths Chapter 8 Heron’s Formula Ex 8.2 help you. If you have any query regarding Karnataka Board Class 9 Maths Chapter 8 Heron’s Formula Exercise 8.2, drop a comment below and we will get back to you at the earliest.

Karnataka Board Class 9 Maths Chapter 12 Circles Ex 12.1

Question 1.
Fill in the blanks : (Answers are filled)
(i) The centre of a circle lies in the interior of the circle, (exterior/interior)
(ii) A point, whose distance from the centre of a circle is greater than its radius lies in the exterior of the circle, (exterior/interior)
(iii) The longest chord of a circle is a diameter of the circle.
(iv) An arc is a semicircle when its ends are the ends of a diameter.
(v) Segment of a circle is the region between an arc and chord of the circle.
(vi) A circle divides the plane, on which it lies, in three parts.

2. Write True or False: Give reasons for your answers.
Question 1.
Line segment joining the centre to any point on the circle is a radius of the circle.
Answer:
True. Each point in the circumference are equidistant from the centre of the circle. That distance is called radius of the circle.

Question 2.
A circle has only finite number of equal chords.
Answer:
False. A circle has infinite number of equal chords because in a circle there are infinite points.

Question 3.
If a circle is divided into three equal arcs, each is a major arc.
Answer:
False. If equilateral triangle is constructed in interior of the circle, the three equal parts are called minor arcs.

Question 4.
A chord of a circle, which is twice as long as its radius, is a diameter of the circle.
Answer:
True. Diameter is a line segment which passes through centre of the circle.

Question 5.
Sector is the region between the chord and its corresponding arc.
Answer:
False. The part which joins centre of the circle to the end points of arc, area coveted with two radii and arc s is called sector.

Question 6.
A circle is a plane figure.
Answer:
True. A circle has two dimension.

KSEEB Solutions for Class 9 Maths

Karnataka Board Class 9 Maths Chapter 11 Areas of Parallelograms and Triangles Ex 11.4

Question 1.
Parallelogram ABCD and rectangle ABEF are on the same base AB and have equal areas. Show that the perimeter of the parallelogram is greater than that of the rectangle.

Solution:
Data: Parallelogram ABCD and rectangle ABEF are on the same base AB and have equal areas.
To Prove: Perimeter of the parallelogram is greater than that of the rectangle.
Proof: ||^grm ABCD and rectangle ABEF are on the same base AB and in between AB || FC.
AB = CD (Opposite sides of ||^grm)
AB = EF (Opposite side of rectangle)
∴ CD = EF
∴ AB + CD = AB + EF …………. (i)
Perpendicular drawn from the vertex to base is only smaller than the remaining lines.
∴ FA < AD and BE < BC
∴ AF + BE < AD + BC ………. (ii)
Comparing (i) and (ii),
(AB + CD + AF + BE) < (AB + EF + AD + BC)
∴ Perimeter of the parallelogram is greater than that of the rectangle.

Question 2.
In Fig., D and E are two points on BC such that BD = DE = EC. Show that ar.(ABD) = ar.(ADE) = ar.(AEC).

Can you now answer the question that you have left in the ‘Introduction’ of this chapter, whether the field of Budhia has been actually divided into three parts of equal area?
Solution:
Data : D and E are two points on BC such that BD = DE = EC.
To Prove: ar.(ABD) = ar.(ADE) = ar.(AEC)
Construction: Draw AM ⊥ BC.

[Remark: Note that by taking BD = DE = EC, the triangle ABC is divided into three triangles ABD,- ADE, and AEC of equal areas. In the same way, by dividing BC into ‘n’ equal parts and joining the points of division so obtained to the opposite vertex of BC, you can divide AABC into ‘n’ triangles of equal areas.]

Question 3.
In Fig., ABCD, DCEF and ABFE are parallelograms. Show that ar. (∆ADE) = ar. (∆BCF).

Solution:
Data: ABCD, DCEF, and ABFE are Parallelograms.
To Prove: ar.(∆ADE) = ar.(∆BCF)
Proof: In ||^grm ABCD,
∴ AD = BC (Opposite sides of ||^grm) ………… (i)
Similarly, in DCFE,
DE = CF (Opposite sides of ||^grm) ………..(ii)
Similarly, in ABFE,
AE = BF (Opposite sides of ||^grm) ………… (iii)
In ∆ADE and ∆BCF,
AD = BC ……… (1)
DE = CF ……… (2)
AE = BF ……… (3)
∴ ∆ADE ≅ ∆BCF (SSS postulate)
∴ ar.(∆ADE) = ar.(∆BCF).

Question 4.
In Fig., ABCD is a parallelogram, and BC is produced to a point Q such that AD = CQ. If AQ intersect DC at P, show that ar(∆BPC) = ar(∆DPQ).
(Hint: Join AC).

Solution:
Data: ABCD is a parallelogram and BC is produced to a point Q such that AD = CQ. AQ intersect DC at P,
To Prove: ar.(∆BPC) = ar.(∆DPQ)
Construction: AE is joined.
Proof: ABCD is a parallelogram.
∆APC and ∆BDC are on BC and in between AB || PC.
∴ ∆APC = ∆BPC …………… (i)
In this fig. AD || CQ and AD = CQ (Data)
∴ ∆BCD is a parallelogram.
∆DQC and ∆AQC are on base QC and in between QC || AD.
∴ ∆DQC = ∆AQC
ar.(∆DQC) – ar.(∆PQC) = ar.(∆ACQ) – ar.(∆PQC)
∴ ar.(∆APC) = ar.(∆DPQ) ………….. (ii)
Comparing (i) and (ii),
ar.(∆BPC) = ar.(∆DPQ).

Question 5.
In Fig., ∆ABC and ∆BDE are two equilateral triangles such that D is the mid-point of BC. If AE intersects BC at F, show that

(i) ar.(∆BDE) = \(\frac{1}{4}\) ar.(∆ABC)
(ii) ar.(∆BDE) = \(\frac{1}{2}\) ar.(∆BAE)
(iii) ar.(∆ABC) = 2ar.(∆BEC)
(iv) ar.(∆BFE) = ar.(∆AFD
(v) ar.(∆BFE) = 2ar.(∆FE:
(vi) ar.(∆FED)= \(\frac{1}{8}\) ar.(∆AFC)
[Hint: Join EC and AD. Show that BE || AC and DE || AB, etc.]
Solution:
Data : ∆ABC and ∆BDE are two equilateral triangles such that D is the mid-point of BC. AE intersects BC at F,
To Prove:
(i) ar.(∆BDE) = \(\frac{1}{4}\) ar.(∆ABC)
(ii) ar.(∆BDE) = \(\frac{1}{2}\) ar.(∆BAE)
(iii) ar.(∆ABC) = 2ar.(∆BEC)
(iv) ar.(∆BFE) = ar.(∆AFD
(v) ar.(∆BFE) = 2ar.(∆FE:
(vi) ar.(∆FED)= \(\frac{1}{8}\) ar.(∆AFC)
Construction: BC and AD are joined. G and H are midpoints of AB and AC resply. GH, GD and HD are joined.
Proof: (i) D is the midpoint of BC.
Now, GH || BC OR GH || BD (Data)
GH = \(\frac{1}{2}\) BC OR GH = BD
∴ BDHG is a parallelogram.
∴ ar.(∆GHD) = ar. (∆BDG)
Similarly, GDCH is a parallelogrm.
∴ ar.(∆GHD) = ar.(∆DCH)
AGDH is a parllelogrm.
∴ ar.(∆GHD) = ar.(∆AGH)
∆ABC = ∆BDG + ∆GHD + ∆AGH + ∆HDC
∴ ∆ABC = 4 × ∆BDE
∴ ar.(∆BDE) = \(\frac{1}{4}\) × ar.(∆ABC).

(ii) ∆BDE = ∆AED
(∆BDE) – (∆FED) = (∆AED) – (∆FED)
∴ (∆BEF) = (∆ADF) ………. (i)
(∆ABD) = (∆ABF) + (∆ADF)
(∆ABD) = (∆ABF) + (∆BEF) [from (i)]
∴ (∆ABD) = (∆ABE) …………. (ii)
But, (∆ABD) = \(\frac{1}{2}\) × (∆ABC)
(∆ABD) = \(\frac{1}{2}\) × 4W(∆BDE)
∴ ar.(∆ABD) = 2 × ar.(∆BDE) ………… (iii)
From (i) and (ii),
ar.(∆ABE) = 2 × ar.(∆BDE)
∴ ar.(∆BED) = \(\frac{1}{2}\) × ar.(∆ABE)

(iii) ∆ABE and ∆BEC are on same base BE and in between AC || BE.
∴ (∆ABE) = (∆BEC)
But, ar.(∆ABE) = 2 × ar.(∆BDE) (proved)
∴ 2ar.(∆BDE) = ar.(∆BEC)
But, ar.(∆BDE) = \(\frac{1}{4}\) × ar.(∆ABC)
∴ 2 × \(\frac{1}{4}\) ar.(∆ABC) = ar.(∆BEC)
∴ r.(∆BEC) = \(\frac{1}{2}\) × ar.(∆ABC)
ar.(∆ABC) = 2 × ar.(∆BEC)

(iv) ∆BED and ∆AED are on same base DE and in between AB || DE.
∴ ar. (∆BED) = ar.(∆AED)
ar.(∆BED) – ar.(∆DEF) = ar.(∆AED) – ar.(∆DEF)
ar.(∆BEF) = ar.(∆AFD)

(v) Let height of ∆ABC be ‘a’ and height of DBDE be V.

(vi) ar.(∆AFC)= ar.(∆AFD) + ar.(∆ADC)
= ar. (∆BEF) + \(\frac{1}{2}\) ar.(∆ABC)
= ar.(∆BEF) + \(\frac{1}{2}\) × 4ar.(∆BED)
= ar.(∆BFE) + 2ar.(∆BDE)
But, ar.(∆BED) = ar.(∆BFE) + ar.(∆FED)
= 2 × ar.(∆FED) + ar.(∆FED)
= 3 × ar.(∆FED)
∴ ar.(∆AFC) = 2 × ar.(∆FED) + 2 × 3 ar.(∆FED)
= 2 × ar.(∆FED) + 6 ×ar.(∆FED)
ar.(∆AFC)= 8 × ar.(∆FED)
ar.(∆FED) = \(\frac{1}{8}\) × ar.(∆AFC)

Question 6.
Diagonals AC and BD of a quadrilateral ABCD intersect each other at P. Show that
ar.(∆APB) × ar.(∆CPD) = ar.(∆APD) × ar.(∆BPC)
[Hint: From A and C, draw perpendiculars to BD).

Solution:
Data: Diagonals AC and BD of a quadrilateral ABCD intersect each other at P.
To Prove: ar.(∆APB) × ar.(∆CPD) = ar.(∆APD) × ar.(∆BPC)
Construction: Draw AM ⊥ DB, CN ⊥ DB.
Proof: ar.(∆APB) × ar.(∆CPD) =

From (i) and (ii), .
ar.(∆APB) × ar.(∆CPD) = ar.(∆APD) × ar.(∆BPC).

Question 7.
P and Q are respectively the mid-points of sides AB and BC of a triangle AABC and R is the mid-point of AP, Show that

(i) ar.(∆PRQ) = \(\frac{1}{2}\) ar.(∆ARC)
(ii) ar.(∆RQC) = \(\frac{3}{8}\) ar.(∆ABC)
(iii) ar.(∆PBQ) = ar.(∆ARC).
Solution:
Data : P and Q are respectively the mid-points of sides AB and BC of a triangle ∆ABC and R is the mid-point of AP.
To Prove:
(i) ar.(∆PRQ) = \(\frac{1}{2}\) ar.(∆ARC)
(ii) ar.(∆RQC) = \(\frac{3}{8}\) ar.(∆ABC)
(iii) ar.(∆PBQ) = ar.(∆ARC).
Proof: P and Q are respectively the mid points of sides aB and BD
∴ PQ || AC and PQ = \(\frac{1}{2}\) AC
(∵ Mid-point theorem)
Mark S such that it is the mid point of AC and join QS.
∴ PQSA is a parallelogram. Its diagonals bisect quadrilateral into two congruent triangles.
ar.(∆PAS) = ar.(∆PQS) = ar.(∆PQA) = ar.(∆QAS)
Similarly, PSCQ, QSCT and PSQB are parallelograms.
∴ ar. (∆PSQ) = ar.(∆CQS)
ar.(∆QSC)= ar.(∆CTQ)
ar. (∆PSQ)=ar. (∆QBP)
∴ ∆PAS = ∆SQP = ∆PAQ = ∆SQA = ∆QSC = ∆CTQ = ∆QBP …………… (i)
and ∆ABC = ∆PBQ + ∆PAS + ∆PQS + ∆QSC
∆ABC = ∆PBQ + ∆PBQ + ∆PBQ + ∆PBQ
∆ABC = 4 × ∆PBQ
ar.(∆ABC) = 4 × ar.(∆PBQ)
ar.(∆PBQ) = \(\frac{1}{4}\) × ar.(∆ABC) …………….. (ii)

(i) P and C are joined.
QR is the median of ∆PAQ.
∴ ar.(∆PQR) = ar.(∆AQR)

Question 8.
In Fig., ABC is a right triangle right angled at A, BCED, ACFG and ABMN are squares on the sides BC, CA and AB respectively. Line segment AX ⊥ DE meets BC at Y. Show that :
(i) ∆MBC ≅ ∆ABD
(ii) ar.(BYXD) = 2ar.(∆MBC)
(iii) ar.(BYXD) = ar.(ABMN)
(iv) ∆FCB ≅ ∆ACE
(v) ar.(CYXE) = 2ar.(∆FCB)

(vi) ar.(CYXE) = ar,(ACFG)
(vii) ar.(BCED) = ar.(ABMN) + ar.(ACFG).
Solution:
Data: ABC is a right-angled triangled at A, BCED, ACFG and ABMN are squares on the sides BC, CA and AB respectively. Line segment AX ⊥ DE meets BC at Y.

To Prove:
(i) ∆MBC ≅ ∆ABD
(ii) ar.(BYXD) = 2ar.(∆MBC)
(iii) ar.(BYXD) = ar.(ABMN)
(iv) ∆FCB ≅ ∆ACE
(v) ar.(CYXE) = 2ar.(∆FCB)
(vi) ar.(CYXE) = ar.(ACFG)
(vii) ar.(BCED) = ar.(ABMN) + ar.(ACFG).
Proof: (i) Each angle of the square is 90°.
∴ ∠ABM = ∠DBC = 90°
∠ABM + ∠ABC = ∠DBC + ∠ABC
∴ ∠MBC = ∠ABD
In ∆MBC and ∆ABD,
∠MBC = ∠ABD (Proved)
MB = AB
BC = AD
∴ ∆MBC ≅ ∆ABD (ASA Postulate)

(ii) ∆MBC = ∆ABD
∴ ar.(∆MBC) = ar.(∆ABD) ………… (i)
AX ⊥ DE, BD ⊥ DE
∴ BD || AX
∆ABD and quadrilateral BYXD are on base BD and in between BD || AX.
ar.(∆ABD) = \(\frac{1}{2}\) × ar.(BYXD)
BYXD = 2 × ar.(∆ABC)
BYXD = 2 × ar.(∆MBC) …………(ii)

(iii) ∆MBC, quadrilateral ABMN are on base MB and in between MB || NC.
∴ ar.(∆MBC) = \(\frac{1}{2}\) × ar.(ABMN)
2 × ar.(∆MBC) = ar.(ABMN)
ar.(BYXD) = ar.(ABMN) ……………….. (iii)

(iv) Each angle of square is 90°.
∴ ∠FCA = ∠BCE = 90°
∠FCA + ∠ACB = ∠BCE + ∠ACB
∴ ∠FCB = ∠ACE
In ∆FCB and ∆ACE,
∠FCB = ∠ACE
FC = AC
CB = CE
∴ ∆FCB ≅ ∆ACE (SAS postulte)

(v) AX ⊥ DE, CE ⊥ DE (Data)
∴ CE || AX
∆ACE, quadrilateral CYXE are on base CE and in between CE || AX.
∴ ar.(∆ACE) = \(\frac{1}{2}\) ar.(CYXE)
ar.(CYXE) = 2 × ∆ACE …………… (iv)
∆FCB ≅ ∆ACE Proved.
∴ W(∆FCB) = W(∆ACE) …………… (v)
Comparing (iv) and (v),
ar.(CYXE) = 2 × ar.(∆FCB) ……………. (vi)

(vi) ∆FCB, and quadrilateral ACFG are on base CF and in between CF || BG.
∴ ar.(∆FCB) = \(\frac{1}{2}\) × ar.(∆CFG)
ar.(ACFG) = 2 × ar.(∆FCB) …………… (vii)

(vii) ar.(BCED) = ar.(BYXD) + ar.(CYXE)
∴ ar.(BCED) = ar.(ABMN) + ar.(ACFG).
[Note: Result (vii) is the famous Theorem of Pythagoras. You shall learn a simpler proof of this theorem in Class X.]

KSEEB Solutions for Class 9 Maths

KSEEB Solutions for Class 9 Maths Chapter 8 Heron’s Formula Ex 8.1 are part of KSEEB Solutions for Class 9 Maths. Here we have given Karnataka Board Class 9 Maths Chapter 8 Heron’s Formula Exercise 8.1.

Karnataka Board Class 9 Maths Chapter 8 Heron’s Formula Ex 8.1

Question 1.
A traffic signal board, indicating ‘SCHOOL AHEAD’, is an equilateral triangle with side ‘a’. Find the area of the signal board, using Heron’s formula. If its perimeter is 180 cm, what will be the area of the signal board ?
Solution:
Let each side of the signal board be ‘a’ unit.

Its’ perimeter,
AB + BC + CA = 180 cm
Area = ?
Perimeter of ∆ABC
a + a + a = 180
∴ 3a = 180
∴ a = \(\frac{180}{3}\)
∴ a = 60 cm
a = b = c = 60 cm
\(s=\frac{a+b+c}{2}=\frac{180}{2}=90\)
Area of ∆ABC A,

Question 2.
The triangular side walls of a flyover have been used for advertisements. The sides of the walls are 122 m, 22 m and 120 m. The advertisements yield an earning of Rs. 5000 per m² per year. A company hired one of its walls for 3 months. How much rent did it pay?

Solution:
Sides of triangle, a = 122m, b = 22m, c = 120m.

Rent for each year, each sq.m. = Rs. 5,000
Rent for each month, each sq.m. = \(\frac{5000}{12}\)
Rent for each quarterly, each sq.m. = 3 × \(\frac{5000}{12}\)
= Rs. 1250
∴ Rent for 1320 sq.m.= 1250 × 1320 = Rs. 1650000
∴ Amount company has to pay is Rs. 1650000.

Question 3.
There is a slide in a park. One of its side walls has been painted in some colour with a message “KEEP THE PARK GREEN AND CLEAN”. If the sides of the wall are 15m, 11m and 6 m, find,the area painted in colour.

Solution:
Sides, a = 15 m, b = 11 m, c = 6 m.

Question 4.
Find the area of a triangle two sides of which are 18 cm and 10 cm and the perimeter is 42 cm.
Solution:
Three sides of triangle be
Let a = 18 cm, b = 10 cm, c = x cm.
Perimeter = 42 cm.
a + b + c = 42
18 + 10 + x = 42
28 + x = 42
∴ x = 42 – 28
∴ x = 14 cm
∴ c = 14 cm

Question 5.
Sides of a triangle are in the ratio of 12: 17: 25 and its perimeter is 540 cm. Find its area.
Solution:
Let sides of triangle be
a = 12x, b = 17x and c = 25x
∴ Perimeter = 12x + 17x + 25x = 540
54x = 540
∴ x = \(\frac{540}{54}\)
∴ x = 10 cm.
∴ a = 12x = 12 × 10 = 120 cm.
b = 17x = 17 × 10 = 170 cm.
c = 25x = 25 × 10 = 250 cm.

Question 6.
An isosceles triangle has perimeter 30 cm and each of the equal sides is 12 cm. Find the area of the triangle.
Solution:
Let the sides of isosceles triangle be a and b.
∴ a = b = 12 cm.
Let the third side be ‘c’.
∴ Perimeter = a + b + c = 30
= 12 + 12 + c = 30
24 + c = 30
c = 30 – 24
c = 6 cm.

We hope the KSEEB Solutions for Class 9 Maths Chapter 8 Heron’s Formula Ex 8.1 help you. If you have any query regarding Karnataka Board Class 9 Maths Chapter 8 Heron’s Formula Exercise 8.1, drop a comment below and we will get back to you at the earliest.

KSEEB Solutions for Class 9 Maths Chapter 7 Quadrilaterals Ex 7.2 are part of KSEEB Solutions for Class 9 Maths. Here we have given Karnataka Board Class 9 Maths Chapter 7 Quadrilaterals Exercise 7.2.

Karnataka Board Class 9 Maths Chapter 7 Quadrilaterals Ex 7.2

Question 1.
ABCD is a quadrilateral in which P, Q, R and S are mid-points of the sides AB, BC, CD and DA. AC is a diagonal. Show that :

(i) SR || AC and SR = \(\frac{1}{2}\) AC
(ii) PQ = SR
(iii) PQRS is a parallelogram.
Solution:
Data: ABCD is a quadrilateral in which P, Q, R, and S are mid-points of the sides AB, BC, CD and DA. AC is a diagonal.
To Prove: (i) SR || AC and SR = \(\frac{1}{2}\) AC
(ii) PQ = SR
(iii) PQRS is a parallelogram.
Proof: (i) In ∆ADC, S, and R are midpoints of AD and DC sides.
As per mid-point theorem,
SR || AC
and SR = \(\frac{1}{2}\) AC.

(ii) In ∆ABC, P and Q are mid-points of AB and BC.
As per mid-point theorem,
PQ || AC
and PQ = \(\frac{1}{2}\) AC
But, SR = \(\frac{1}{2}\) AC (Proved)
∴ PQ = SR

(iii) PQ = SR (Proved)
SR || AC and PQ || AC
∴ SR || PQ
Opposite sides of a quadrilateral PQRS are equal and parallel, hence PQRS is a parallelogram.

Question 2.
ABCD is a rhombus and P, Q, R, and S are the mid-points of the sides AB, BC, CD, and DA respectively. Show that the quadrilateral PQRS is a rectangle.

Solution:
Data: ABCD is a rhombus and P, Q, R, and S are the mid-points of the sides AB, BC, CD, and DA respectively.
To Prove: PQRS is a rectangle.
Construction: Diagonals AC and BD are drawn.
Proof: To prove PQRS is a rectangle, one of its angles should be the right angle.
In ∆ADC, S and R are the midpoints of AD and DC.
∴ SR || AC
SR = \(\frac{1}{2}\) AC (mid-point formula)
In ∆ABC, P and Q are the midpoints AB and BC.
∴ PQ || AC
PQ = \(\frac{1}{2}\) AC.
∴ SR || PQ and SR = PQ
∴ PQRS is a parallelogram.
But diagonals of a rhombus bisect at right angles. 90° angle is formed at ’O’.
∴ ∠P = 90°
∴ PQRS is a parallelogram, each of its angles is a right angle.
This is the property of the rectangle.
∴ PQRS is a rectangle.

Question 3.
ABCD is a rectangle and P, Q, R, and S are mid-points of the sides AB, BC, CD, and DA respectively. Show that the quadrilateral PQRS is a rhombus.

Solution:
Data: ABCD is a rectangle and P, Q, R, and S are mid-points of the sides AB, BC, CD, and DA respectively.
To Prove: PQRS is a rhombus.
Construction: Diagonals AC and BD are drawn.
Proof: In ∆ABC, P and Q are the mid-points of AD and BC.
∴ PQ || AC (Mid-point theorem)
PQ = \(\frac{1}{2}\) AC ………….. (i)
Similarly, in ∆ADC, S and R are the mid-points of AD and CD.
∴ SR || AC
SR = \(\frac{1}{2}\) AC …………… (ii)
Similarly, in ∆ABD,
SP || BD
SP = \(\frac{1}{2}\) BD ……………….. (iii)
Similarly, in ∆BCD,
QR || BD
QR = \(\frac{1}{2}\) BD ……………… (iv)
From (i), (ii), (iii) and (iv),
PQ = QR = SR = PS and Opposite sides are parallel.
∴ PQRS is a rhombus.

Question 4.
ABCD is a trapezium in which AB || DC, BD is a diagonal and E is the mid-point of AD. A line is drawn through E parallel to AB intersecting BC at F. Show that F is the mid-point of BC.

Solution:
Data: ABCD is a trapezium in which AB || DC, BD is a diagonal, and E is the mid-point of AD.
A line is drawn through E parallel to AB intersecting BC at F.
To Prove: F is the mid-point of BC.
Proof: Straight line EF cuts at G.
In ∆ABD, E is the mid-point of AD (Data)
EF || AB
EG || AB
∴ G is the mid-point of BD.
∵ Converse of mid-point formula.
DC || AB and EF || AB
⇒ DC || EF
In ∆BDC,
GF || DC
G is the mid-point of BD (proved)
∴ F is the mid-point of BC.

Question 5.
In a parallelogram ABCD, E and F are the mid-points of sides AB and CD respectively. Show that the line segments AF and EC trisect the diagonal BD.

Solution:
Data: In a parallelogram ABCD, E and F are the mid-points of sides AB and CD respectively.
To Prove: Line segments AF and EC trisect the diagonal BD.
Proof: ABCD is a parallelogram.
AB || DC and AB = DC
\(\frac{1}{2}\) AB = \(\frac{1}{2}\) DC
AE = CF
and AE || CF (∵ AB || CD)
In AECF quadrilateral,
∴ AE || CF and AE = CF.
∴ AECF is a parallelogram.
∴ AF || EC
In ∆DQC, PF || QC (∵ AF || EC)
∴ P is the mid-point of DQ.
∴ DP = PQ ………….. (i)
In ∆APB, EQ || AP.
But E is the mid-point of AB (Data)
∴ Q is the mid-point of PB.
∴ PQ = QB …………… (ii)
From (i) and (ii),
DP = PQ = QB
∴ AF and EC line segments trisect the diagonal BD.

Question 6.
Show that the line segments joining the mid-points of the opposite sides of a quadrilateral bisect each other.

Solution:
Data : P, Q, R and S are mid-points of AB, BC, CD and DA respectively in quadrilateral ABCD.
To Prove: PR and SQ line segments bisect mutually.
Construction: Join the diagonal AC.
Proof: In ∆ADC,
S and R are the mid-points of AD and DC.
∴ SR || AC
SR = \(\frac{1}{2}\) AC ………… (i) (Mid-point Theorem)
Similarly, in ∆ABC,
PQ || AC
PQ = \(\frac{1}{2}\) AC …………. (ii)
From (i) and (ii),
SR = PQ and SR || PQ
∴ PQRS is a parallelogram.
PR and SQ are the diagonals of parallelogram PQRS.
∴PR and SQ meet at O’.

Question 7.
ABC is a triangle right angled at C. A line through the mid-point M of hypotenuse AB and parallel to BC intersects AC at D. Show that

(i) D is the mid-point of AC.
(ii) MD ⊥ AC
(iii) CM = MA = \(\frac{1}{2}\)AB.
Solution:
Data: ABC is a triangle right angled at C. A line through the mid-point M of hypotenuse AB and parallel to BC intersects AC at D.
To Prove: (i) D is the mid-point of AC.
(ii) MD ⊥ AC
(iii) CM = MA = \(\frac{1}{2}\) AB.
Proof: (i) In ∆ABC,
M is the mid point of AB and
MB || BC Mid-point of BC (Data)
∴ D is the mid-point of AC (Mid-point formula)

(ii) MD || BC
∠BCD + ∠MDC = 180° (Sum of interior angles)
90 + ∠MDC = 180°
∠MDC = 180 – 90
∠MDC = 90°
∴ MD ⊥ AC

(iii) In this fig. CM is joined.
M is the mid-point of AB.

∴ MA = \(\frac{1}{2}\) AB ……….. (i)
Now, in ∆AMD and ∆CMD.
AD = DC (proved)
∠MDC = ∠MDA = 90°
MD is common.
∴∆AMD ≅ ∆CMD (SAS Postulate)
∴ CM = MA …………… (ii)
Comparing (i) and (ii),
CM = MA = \(\frac{1}{2}\) AB.

We hope the KSEEB Solutions for Class 9 Maths Chapter 7 Quadrilaterals Ex 7.2 helps you. If you have any query regarding Karnataka Board Class 9 Maths Chapter 7 Quadrilaterals Exercise 7.2, drop a comment below and we will get back to you at the earliest.

Karnataka Board Class 9 Maths Chapter 11 Areas of Parallelograms and Triangles Ex 11.3

Question 1.
In Fig., E is any point on median AD of a ∆ABC. Show that ar. (ABE) = ar.(ACE).

Solution:
Data: E is any point on Median AD of an ∆ABC.
To Prove: ar.(∆ABE) = ar. (∆ACE)
Proof: In ∆ABC, AD is the median.
∴ ∆ABD = ∆ACD ……….. (i)
In ∆EBC, DE is the median.
∴ ∆EBC = ∆ECD …………. (ii)
Subtracting (ii) from (i),
∆ABD – ∆EBD = ∆ACD – ∆ECD
∴ ar. (∆ABE) = ar. (∆ACE).

Question 2.
In a triangle ABC, E is the mid-point of median AD. Show that ar.(∆BED) = \(\frac{1}{4}\) ar.(∆ABC).

Solution:
Data: In a triangle ABC, E is the midpoint of median AD.
To Prove: ar.(∆BED) = \(\frac{1}{4}\) ar.(∆ABC)
Proof: D is the midpoint of BD.
∴ BD = \(\frac{1}{2}\) BC ……… (i)
E is the midpoint of AD.
∴ DE = \(\frac{1}{2}\) AD …………(ii)

Question 3.
Show that the diagonals of a parallelogram divide it into four triangles of equal area.

Solution:
Data: Diagonals of ABCD, AC and BD intersect at ‘O’.
To Prove: ar.(∆AOD) = ar.(∆DOC) = ar.(∆COB) = ar.(∆AOB)
Proof: ‘O’ is the mid-point of diagonals AC and BD.
DO is the median of ∆ADC.
∴ ar.(∆AQD) = ar.(∆DOC) ……….. (i)
CO is the median of ∆DCB.
∴ ar. (∆DOC) = ar.(∆BOC) ……….. (ii)
BO is the median of ∆ABC.
∴ ar.(∆BOC) = ar.(∆AOB) …………. (iii)
AO is the median of ∆ADB.
∴ ar.(∆AOB) = ar.(∆AOD) …………. (iv)
From (i), (ii), (iii) and (iv),
ar. (∆AOD) = ar. (∆DOC) = ar. (∆COB) = ar. (∆AOB)
∴ The diagonals of a parallelogram divide it into four triangles of equal area.

Question 4.
In Fig., ABC and ABD are two triangles on the same base AB. If line- segment CD is bisected by AB at O, show that ar.(∆ABC) = ar.(∆ABD).

Solution:
Data ∆ABC and ∆ABD are two triangles on the same base AB.
Line segment CD bisected by AB at O’.
To Prove: ar.(∆ABC) = ar.(∆ABD)
Prove: AB bisects line segment CD at O’.
∴ OC = OD
In ∆ADC, AO is the median on CD.
∴ ar.(∆AOD) = ar.(∆AOC) …………… (i)
In ∆BDC, BO is the median on CD.
∴ ar.(∆BOD) = ar.(∆BOC) ……………. (ii)
Adding (i) and (ii),
ar.(∆AOD) + ar.(∆BOD) = ar.(∆AOC)+ ar.(∆BOC)
∴ ar.(∆ABD) = ar.(∆ABC)

Question 5.
D, E, and F are respectively the midpoints of the sides BC, CA, and AB of an ∆ABC. Show that

(i) BDEF is a parallelogram.
(ii) ar.(∆DEF) = \(\frac{1}{4}\) ar. (∆ABC)
(iii) ar.(BDEF) = \(\frac{1}{2}\) ar.(∆ABC).
Solution:
Data : D, E, and F are respectively the mid-points of the sides BC, CA, and AB of an ∆ABC.
To Prove:
(i) BDEF is a parallelogram.
(ii) ar.(ADEF) = \(\frac{1}{4}\) ar.(∆ABC)
(iii) ar.(BDEF) = \(\frac{1}{2}\) ar.(∆ABC).
Proof: (i) In ∆ABC, D, E and F are respectively mid-points of the sides BC, CA and AB.
∴ BD = \(\frac{1}{2}\) BC
and EF = \(\frac{1}{2}\) BC (mi-point theorem) z .
EF || BC
∴ BD = EF
BD || EF
∴ BDEF is a parallelogram.

(ii) Diagonals of BDEF divides into two congruent triangles.
∴ ∆BDF ≅ ∆DEF ………… (i)
Similarly, in DCEF,
∆DCE ≅ ∆DEF …………. (ii)
Similarly, in AEDF,
∴ ∆AEF ≅ ∆DEF ……….. (iii)
From (i), (ii) and (iii),
ar.(∆BDF) = ar.(∆DEF) = ar.(∆DCE) = ar.(∆AEF)
∴ 4 × ar.(∆DEF) = ar.(∆ABC)
∴ ar.(∆DEF) = \(\frac{1}{4} \)ar.(∆ABC)

(iii) ar.(∆BED) + ar.(∆DCE) + ar.(∆AEF) + ar.(∆DEF) = ar.(∆ABC)
∆BDF + ∆BDF + ∆DEF + ∆DEF = ar.(∆ABC)
2 × ∆BDE + 2 × ∆DEF = ar.(∆ABC)
2(∆BDF + ∆DEF) = ar.(∆ABC)
2 × ar. (BDEF) = ar.(∆ABC)
ar.(BDEF) = \(\frac{1}{2}\) × ar.(∆ABC)

Question 6.
In Fig., diagonals AC and BD of quadrilateral ABCD intersect at O such that OB = OD. If AB = CD, then show that:

(i) ar.(∆DOC) = ar.(∆AOB)
(ii) ar.(∆DCB) = ar.(∆ACB)
(iii) DA || CB or ABCD is a parallelogram.
[Hint: From D and B, draw perpendiculars to AC.]
Solution:
Data: Diagonals AC and BD of quadrilateral ABCD intersect at O such that OB = OD. If AB = CD,
To Prove:
(i) ar.(∆DOC) = ar.(∆AOB)
(ii) ar.(∆DCB) = ar.(∆ACB)
(iii) DA || CB or ABCD is a parallelogram

Construction: Draw DM ⊥ AC and BN ⊥ AC.
Proof: (i) DM ⊥ CO, BN ⊥ AO.
In ∆DOM and ∆BON,
DO = BO (data)
∠DMO = ∠BNO = 90° (Construction)
∠DOM = ∠BON (Vertically opposite angles)
∴ ∆DOM ≅ ∆BON (AAS Postulate)
∴ DM = BN
∴ ar.(∆DOM) = ar.(∆BON) ………… (i)
In ∆DMC, and ∆BNA,
∠DMC = ∠BNA = 90°
DC = AB (Data)
DM = BN (Proved)
∴ ⊥∆DMC = ⊥∆BNA
∴ ar. (∆DMC) = ar, (∆BNA) ………… (ii)
Adding (i) and (ii),
ar.(∆DOM)+ ar.(∆DMC) = ar.(∆BON) + ar.(∆BNA)
∴ (∆DOC) = (∆AOB) …………….. (iii)

(ii) (∆DOC) = (∆AOB) (Proved)
Adding ∆BOC to both sides,
∆DOC + ∆BOC = ∆AOB + ∆BOC
∴ ∆DCB = ∆ACB.

(iii) From equation (ii), ∆DMC = ∠BNA
∴ ∠DCM = ∠BAN.
These are a pair of alternate angles.
∴ AB || DC
and also AB = DC (Data)
∴ ABCD is a parallelogram OR DA || CB.

Question 7.
D and E are points on sides AB and AC respectively of ∆ABC such that ar.(∆DBC) = ar.(∆EBC). Prove that, DE || BC.

Solution:
Data : D and E are mints on sides AB and AC respectively of ∆ABC such that ar.(∆DBC) = ar.(∆EBC).
To Prove DE || BC.
Construction: BE and DC joined.
Proof: ∆DBC and ∆EBC are on base BC.
ar.(∆DBC) = ar. (∆EBC) (Data)
Two triangles on the same base (or equal bases) and between the same parallel are equal in area.
∴ BC || DE.

Question 8.
XY is a line parallel to side BC of a triangle ABC. If BE || AC and CF || AB meet XY at E and F respectively, show that ar.(∆ABE) = ar.(∆ACF).

Solution:
Data: XY is a line parallel to side BC of a triangle ABC.
If BE || AC and CF || AB meet XY at E and F respectively.
To Prove: ar.(∆ABE) = ar.(∆ACF).
Proof: XY || BC
∴ EY || BC
and BE || AC
BE || YC
∴ BEYC is a parallelogram.
Similarly, BCFX is a parallelogram.
BEYC and BCFX are on base BC and between BC || EF.
∴ ar.(BEYC) = ar.(BCFX) ………….. (i)
Now, ∆ABE and BEYC are on base BE and in between BE || AC.
∴ ar.(∆ABE) = \(\frac{1}{2}\) ar.(BCYX) …………… (ii)
Similarly, ∆ACF and BCFX are on base CF and in between CF || AB.
∴ ar.(∆ACF) = \(\frac{1}{2}\) ar.(imggrectBCFX) …………….. (iii)
Comparing (i), (ii) and (iii),
∴ ar. (∆ABE) = ar. (∆ACF)

Question 9.
The side AB of a parallelogram ABCD is produced to any point P. A line through A and parallel to CP meets CB produced at Q and then parallelogram PBQR is completed. Show that ar.(ABCD) = ar(PBQR).
[Hint: Join AC and PQ. Now compare ar.(∆ACQ) and ar.(∆APQ)]
Solution:
Data: The side AB of a parallelogram ABCD is produced to any point P. A line through A and parallel to CP meets CB produced at Q and then parallelogram PBQR is completed.

To Prove: ar.(ABCD) = ar.(PBQR)
Construction: AC and CQ are joined.

Proof: ∆ACQ and ∆APQ are on base AQ and in between AQ || CP.
∴ (∆ACQ) = (∆APQ)
Now, ∆ACQ – ∆ABQ = ∆APQ – ∆ABQ
∴ ∆ABC = ∆QBP ………….. (i)
Diagonals divides quadrilateral into two congruent triangles.

Question 10.
Diagonals AC and BD of a trapezium ABCD with AB || DC, intersect each other at ‘O’. Prove that ar.(∆AOD) = ar.(∆BOC).

Solution:
Data: Diagonals AC and BD of a trapezium ABCD with AB || DC, intersect each other at ‘O’.
To Prove: ar.(∆AOD) = ar.(∆BOC)
Proof: ∆ABD and ∆ABC are on base AB and between AB || DC.
∴ ar. (∆ABD) = ar. (∆ABC)
Subtracting ar.(∆ABO) on both sides,
ar.(∆ABD) – ar.(∆ABO) = ar.(∆ABC) – ar.(∆ABO)
∴ ar.(∆AOD) = ar.(∆BOC).

Question 11.
In fig., ABCDE is a pentagon. A line through B parallel to AC meets DC produced at F. Show that

(i) ar.(∆ACB) = ar.(∆ACF)
(ii) ar.(∆EDF) = ar.(ABCDE).
Solution:
Data: ABCDE is a pentagon. A line through B parallel to AC meets DC produced at F.
To Prove: (i) ar.(∆ACB) = ar.(∆ACF)
(ii) ar.(AEDF) = ar.(ABCDE)
Proof: (i) ∆ACB and ∆HCF are on base AC and between AC || BF.
∴ ar.(∆ACB) = ar.(∆ACF)
(ii) Now, W(∆ACB) = W(∆ACF) proved,
adding ACDE on both sides,
ar.(∆ACB) + ar.(ACDE) = ar.(∆ACF) + ar.(ACDE)
∴ ar.(ABCDE) = ar.(AEDF)

Question 12.
A villager Itwaari has a plot of land of the shape of a quadrilateral. The Gram Panchayat of the village decided to take over some portion of his plot from one of the corners to construct a Health Centre. Itwaari agrees to the above proposal with the condition that he should be given an equal amount of land in lieu of his land adjoining his plot so as to form a triangular plot. Explain how this proposal will be implemented.

Solution:
Let the shape of plot be quadrilateral ABCD. Itwaari agrees to give land as follows:
Diagonal AC of a quadrilateral is joined. BA side of land ABCD is produced to E.
CE and DE || AC are drawn.
Now, ∆ACD and ∆ACE are on same base aC and in between AC || DE.
∴ ar.(∆ACD) = ar.(∆ACE)
Subtracting ar.(∆AOC) on both sides
ar (∆ACD) – ar(∆AOC) = ar(∆ACE) – ar(∆AOC)
ar.(∆DOC) = ar.(∆AOC).
Now, Itwaari agrees to give ADOC to construct a Health Centre. Area of ∆AOE and ∆CEB is remains with Itrwaari.

Question 13.
ABCD is a trapezium with AB || DC. A line parallel to AC intersects AB at X and BC at Y. Prove that area(ADx) = area(ACy). [Hint: Join Cx.)

Solution:
Data : ABCD is a trapezium with AB || DC. A line parallel to AC intersects AB at x and BC at y.
To Prove: ar.(ADx) = ar.(ACy)
Construction: Join CX.
Proof: ABCD is a trapezium.
∴ AB || DC
∆ADx and ∆ACx are on same base Ax and in between AB || DC.
∴ (∆ADx) = (∆ACx) ………… (i)
But, AC || xy (Data)
∆ACx and ∆ACy are on same base AC and in between AC || xy.
∴ (∆ACx) = (∆ACy) …………. (ii)
From comparing (i) and (ii),
∴ (∆ADx) = (∆ACy).

Question 14.
In Fig., AP || BQ || CR.
Prove that ar.(AQC) = ar.(PBR).

Solution:
Data: AP || BQ || CR.
To Prove: ar.(∆AQC) = ar.(∆PBR)
Proof: AP || BQ. CR is a line segment.
∆ABQ and ∆PBQ are on base BQ and in between AP || BQ
ar.(∆ABQ) = ar.(∆PBQ) ………….. (i)
∆BQC and ∆BQR are on base BQ and in between BQ || CR.
∴ ar.(∆BQC) = ar.(∆BQR) ………….. (ii)
Adding (i) and (ii),
ar.(∆ABQ) + ar.(∆BQC) = ar.(∆PBQ) + ar.(∆BQR)
∴ ar.(∆AQC) = ar.(∆PBR).

Question 15.
Diagonals AC and BD of a quadrilateral ABCD intersect at O in such a way that ar.(AOD) = ar.(BOC). Prove that ABCD is a trapezium.

Solution:
Data: Diagonals AC and BD of a quadrilateral ABCD intersect at O in such a way that ar.(AOD) = ar.(BOC).
To Prove: ABCD is a trapezium.
Proof: ar.(∆AOD) = ar.(∆BOC) (Data)
Adding ar.(∆ODC) on both sides,
ar.(∆AOD) + ar.(∆ODC) = ar.(∆BOC) + ar.(∆ODC)
∴ ar.(∆ADC) = ar.(∆BDC)
Area of these triangles are equal, they are on base DC and in between DC, AB straight lines.
∴ DC || AB.
Now, one pair of opposite sides of a quadrilateral are parallel, hence ABCD is a trapezium.

Question 16.
In Fig., ar.(DRC) = ar.(DPC) and ar.(BDP) = ar.(ARC). Show that both the quadrilaterals ABCD and CPR are trapeziums.

Solution:
Data: in fig. ar.(DRC) = ar.(DPC) and
ar.(BDP) = ar.(ARC).
To Prove: (i) ABCD is a trapezium
(ii) DCPR is a trapezium.
Proof: (i) ar.(∆BDP) = ar.(∆ARC) (Data)
ar.(∆DPC) + ar.(∆DCB) = ar.(∆DCR) + ar.(∆DCA)
But, ∆DRC = ∆DPC. (Data)
∴ ar.(∆DCB) = ar.(∆DCA)
These are on same base DC and in between straight lines DC and AB.
∴ DC || AB.
Now, one pair of opposite sides of quadrilateral ABCD are parallel, hence ABCD is a trapezium.

(ii) ∆DRC and ∆DPC are on the same base DC and in between DC and RP and they are equal in area.
∴ DC || RP.
∴ Hence ‘Quadrilateral’ DCPR is a trapezium.

KSEEB Solutions for Class 9 Maths

KSEEB Solutions for Class 9 Maths Chapter 7 Quadrilaterals Ex 7.1 are part of KSEEB Solutions for Class 9 Maths. Here we have given Karnataka Board Class 9 Maths Chapter 7 Quadrilaterals Exercise 7.1.

Karnataka Board Class 9 Maths Chapter 7 Quadrilaterals Ex 7.1

Question 1.
The angles of a quadrilateral are in the ratio 3 : 5 : 9 : 13. Find all the angles of the quadrilateral.

Solution:
The angles of a quadrilateral are in the ratio 3 : 5 : 9 : 13.
Let ∠A : ∠B : ∠C : ∠D = 3 : 5 : 9 : 13.
Sum of ratio = 3x + 5x + 9x + 13x = 30x
Sum of 4 angles of quadrilateral is 360°
∴ ∠A + ∠B + ∠C + ∠D = 360°
3x + 5x + 9x + 13x = 360
30x = 360
∴ x = \(\frac{360}{30}\) = 12°
∠A = 3x = 3 × 12 = 36°
∠B = 5x = 5 × 12 = 60°
∠C = 9x = 9 × 12 = 108°
∠D = 13x = 13 × 12 = 156°

Question 2.
If the diagonals of a parallelogram are equal, then show that it is a rectangle.

Solution:
Data: Diagonals of a parallelogram are equal.
To Prove: ABCD is a rectangle.
Proof: Now ABCD is a parallelogram and diagonal AC = Diagonal BD (Data)
In ∆ABC and ∆ABD,
BC = AD (Opposite sides of a quadrilateral)
AC = BD (Data)
AB common.
∴ ∆ABC ≅ ∆ABD (SSS postulate)
∠ABC = ∠BAD
But, ∠ABC + ∠BAD = 180°
∠ABC + ∠ABC = 180°
2 ∠ABC = 180°
∴ ∠ABC = 90°
If an angle of a parallelogram is a right angle, it is called a rectangle.
∴ ABCD is a rectangle.

Question 3.
Show that if the diagonals of a quadrilateral bisect each other at right angles, then it is a rhombus.

Solution:
Data: ABCD is a parallelogram and diagonals AC and BD bisect at right angles at O’.
To Prove: ABCD is a rhombus.
Proof: Here, AC and BD bisect each other at right angles.
∴ AO = OC
BO = OD
and ∠AOB = ∠BOC = ∠COD = ∠AOD = 90°
If sides are equal to each other, then ABCD is said to be a rhombus.
Now, ∆AOD and ∆COD,
AO = OC (Data)
∠AOD = ∠COD = 90° (Data)
OD is common.
∴ ∆AOD ≅ ∆COD (SAS Postulate)
∴ AD = CD …………… (i)
Similarly,
∆AOD = ∆AOB
AD = AB ………… (ii)
∆AOB ≅ ∆COB
∴ AB = BC ……….. (iii)
∆COB ≅ ∆COD
∴ BC = CD ……………. (iv)
From (i), (ii), (iii) and (iv),
AB = BC = CD = AD
All 4 sides of parallelogram ABCD are equal, then it is a rhombus.

Question 4.
Show that the diagonals of a square are equal and bisect each other at right angles.

Solution:
Data: ABCD is a parallelogram. Its diagonals are AC and BD. They meet at ‘O’.
To Prove: i) AO = OC
BO = OD
ii) ∠AOB = ∠BOC = ∠COD = ∠DOA = 90°.
Proof: In ∆ABC and ∆ABD,
BC = AD (diagonals of square is equal)
∠ABC = ∠BAD = 90° (angles of square)
AB is common.
∴ ∆ABC ≅ ∆ABD (SAS Postulate.)
In ∆AOB and ∆COD,
AB = DC (sides of square)
∠OAB = ∠OCD (alternate angles)
∠OBA = ∠ODC (alternate angles)
∴ ∆AOB ≅ ∆COD (ASA Postulate)
AO = OC
BO = OD …………….. (i)
Similarly,
∆AOB ≅ ∆BOC.
∴ ∠AOB = ∠BOC = 90°
Now, ∆COD ≅ ∆AOD, then
∴ ∠COD = ∠DOA = 90°
∴ ∠AOB = ∠BOC = ∠COD = ∠DOA = 90° …………. (ii)
from (i) and (ii)
∴ Sides of a square are equal and bisect at right angles.

Question 5.
Show that if the diagonals of a quadrilateral are equal and bisect each other at right angles, then it is a square.

Solution:
Data: ABCD is a quadrilateral and bisects each other at right angles, then it is a square.
AO = OC and AC = BD
BO = OD
∠AOB = ∠BOC = ∠COD + ∠DOA = 90°.
To Prove: ABCD is a square.
Proof: In ∆AOB and ∆COD,
AO = OC
BO = OD (Data)
∠AOB = ∠COD (Vertically opposite angles)
∴ ∆AOB ≅ ∆COD (SAS Postulate)
AB = CD …………. (i)
∠ABO = ∠CDO
∴ AB || CD ………… (ii)
From (i) and (ii) ABCD is a parallelogram.
Now, in ∆AOD and ∆COD,
AO = OC (Data)
∠AOD = ∠COD = 90° (Data)
OD is common
∴ ∆AOD ≅ ∆COD (SAS Postulate)
AD = CD …………. (iii)
AD = BC ………….. (iv)
From (ii), (iii) and (iv)
AB = BC = CD = AD
Four sides of a quadrilateral are they are equal to each other and bisect each other at right angles, then it is a square.
∴ ABCD is a square.

Question 6.
Diagonal AC of a parallelogram ABCD bisects ∠A. Show that

(i) it bisects ∠C also,
(ii) ABCD is a rhombus.
Solution:
Data: Diagonal AC of a parallelogram ABCD bisects ∠A.
To Prove: (i) Diagonal AC also bisects ∠C.
(ii) ABCD is a rhombus.
Proof: i) Diagonal AC also bisects ∠A.
∴ ∠DAC = ∠BAC …………… (i)
But, ∠DAC= ∠BCA (alternate angles)(ii)
∠BAC=∠DCA (alternate angles)(iii)
From (i), (ii) and (iii),
∠BCA = ∠DCA
∴ AC bisects ∠C.

(ii) ∠DAC = ∠DCA
∴ AD = DC
But, AD = BC
∴ DC = AB
∴ AB = BC = CD = DA
∴ ABCD is a rhombus.

Question 7.
ABCD is a rhombus. Show that diagonal AC bisects ∠A as well as ∠C and diagonal BD bisects ∠B as well as ∠D.

Solution:
ABCD is rhombus. Diagonal AC bisects ∠A and ∠C.
To Prove: Diagonal BD bisects ∠B and ∠D.
Proof: In a rhombus all sides are equal and opposite angles are equal to each other.
In ∆ABC,
AB = BC
∴ ∠BAC = ∠BCA
∠B AC = ∠DCA (alternate angles)
∴ ∠BCA = ∠DCA …………… (i)
∴ AC bisects ∠C.
Now, ∠BCA = ∠DAC (alternate angles)
∠BAC = ∠DAC
∴ AC bisects ∠A
In ∆ABD,
AD = DB
∴ ∠ABD = ∠ADB
∠ABD = ∠CDB
∴ BD bisects ∠D.
∠ADB = ∠CBD
∠ABD = ∠CBD
∴ BD bisects ∠B.

Question 8.
ABCD is a rectangle in which diagonal

(i) ABCD is a square,
(ii) Diagonal BD bisects ∠B as well as ∠D.
Solution:
Data: ABCD is a rectangle in which diagonal AC bisects ∠A as well as ∠C.
To Prove:
(i) ABCD is a square.
(ii) Diagonal BD bisects ∠B as well as ∠D.
Proof: (i) ABCD is a rectangle. AC and BD diagonals intersect at O’.
AB = DC (Opposite sides of rectangle)
AD = BC
∠BAD = ∠BCD (Opposite angles)
∴ \(\frac{1}{2}\) ∠BAD = \(\frac{1}{2}\) ∠BCD
∴ ∠DAC = ∠DCA (AC bisects ∠A)
∴ DC = DA
But, AB = CD and DA = BC (Data)
∴ AB = BC = CD = AD
∴ ABCD is a square.

(ii) In ∆ABD.
AB = AD
∠ABD = ∠ADB
∠ABD = ∠CDB (alternate angles)
∠ADB = ∠CDB
∴ BD bisects ∠D.
∠ADB = ∠CDB (alternate angles)
∠ABD = ∠CBD (alternate angles)
∴ BD bisects ∠B.

Question 9.
In parallelogram ABCD. two points P and Q are taken on diagonal BD such that DP = BQ. Show that:

(i) ∆APD ≅ ∆CQB
(ii) AP = CQ
(iii) ∆AQB ≅ ∆CPD
(iv) AQ = CP
(v) APCQ is a parallelogram.
Solution:
Data: In parallelogram ABCD, two points P and Q are taken on diagonal BD such that DP = BQ.
To Prove:
(i) ∆AAPD ≅ ∆CQB
(ii) AP = CQ
(iii) ∆AQB ≅ ∆CPD
(iv) AQ = CP
(v) APCQ is a parallelogram.
Proof: ABCD is a parallelogram.
(i) In ∆APD and ∆CQB,
AD = BC (opposite sides)
∠ADP = ∠CBQ (alternate angles)
DP = BQ (Data)
∴ ∆APD ≅ ∆CQB (ASA Postulate)

(ii) AP = CQ

(iii) In ∆AQB and ∆CPD,
AB = CD (opposite sides)
∠ABQ = ∠CDP (alternate angles)
BQ = DP (Data)
∴ ∆AQB ≅ ∆CPD (ASA Postulate)

(iv) ∴ AQ = CP

(v) In ∆AQPand ∆CPQ,
AP = CQ (proved)
AQ = PC PQ is common.
∴ ∆AQP ≅ ∆CPQ
∴ ∠APQ = ∠CQP
These are a pair of alternate angles.
∴AP || PC
Similarly, AQ || PC
∴ APCQ is a parallelogram.

Question 10.
ABCD is a parallelogram and AP and CQ are perpendiculars from vertices A and C on diagonal BD. Show that

(i) ∆APB ≅ ∆CQD
(ii) AP = CQ.
Solution:
Data: ABCD is a parallelogram and AP and CQ are perpendiculars from vertices A and C on diagonal BD.
To Prove:
(i) ∆APB ≅ ∆CQD
(ii) AP = CQ.
Proof: ABCD is a parallelogram. BD is diagonal.
AP ⊥ BD and CQ ⊥ BD.

(i) In ∆APB and ∆CQD,
∠APB = ∠CQD = 90°
AB = CD (opposite sides)
∠ABP = ∠CQD (alternate angles)
∴ ∆APB ≅ ∆CQD (AAS postulate)

(ii) ∴ AP = CQ.

Question 11.
In ∆ABC and ∆DEF, AB = DE, AB || DE, BC = EF and BC || EF. Vertices A, B, and C are joined to vertices D. E and F respectively. Show that

(i) Quadrilateral ABED is a parallelogram.
(ii) Quadrilateral BEFC is a parallelogram.
(iii) AD || CF and AD = CF.
(iv) quadrilateral ACFD is a parallelogram.
(v) AC = DF
(vi) ∆ABC ≅ ∆DEF.
Solution:
Data: In ∆ABC and ∆DEF, AB = DE, AB || DE, BC = EF and BC || EF.
Vertices A, B, and C are joined to vertices D, E, and F respectively.
To Prove:
(i) Quadrilateral ABED is a parallelogram.
(ii) Quadrilateral BEFC is a parallelogram.
(iii) AD || CF and AD = CF.
(iv) quadrilateral ACFD is a parallelogram.
(v) AC = DF
(vi) ∆ABC ≅ ∆DEF.
Proof: (i) AB = DE and AB||DE (Data)
∴ BE = AD and BE || AD
∴ ABCD is a parallelogram.

(ii) Similarly, BC = EF and BC || EF.
∴ BE = CF
BE || CF
∴ BEFC is a parallelogram.

(iii) ABED is a parallelogram.
∴ AD = BE
AD || BE ………. (i)
Similarly, BEFC is a parallelogram.
∴ CF = BE
CF || BE ……………. (ii)
Comparing (i) and (ii),
AD = CF and AD || CF,

(iv) In a quadrilateral ACFD,
AD = CF
AD || CF (proved)
∴ AC = DF
AC || DF
∴ ACFD is a parallelogram.

(v) ACFD is a parallelogram.
AC = DF (opposite sides).

(vi) In ∆ABC and ∆DEF,
AB = DE (Data)
BC = EF (opposite sides of a
parallelogram)
AC = DF (Opposite sides of a
parallelogram)
∴ ∆ABC ≅ ∆DEF (SSS postulate).

Question 12.
ABCD is a trapezium in which AB||CD and AD = BC . Show that

(i) ∠A = ∠B
(ii) ∠C = ∠D
(iii) ∆ABC ≅ ∆BAD
(iv) diagonal AC = diagonal BD.
(Hint: Extend AB and draw a line through C parallel to DA intersecting AB produced at E.]
Solution:
Data: ABCD is a trapezium in which AB || CD and AD = BC.
To Prove:
(i) ∠A = ∠B
(ii) ∠C = ∠D
(iii) ∆ABC ≅ ∆BAD
(iv) diagonal AC = diagonal BD.
Construction: Produce straight line AB.
Draw a parallel line DA, parallel to AB such that the produced line meet this at E.
Proof: (i) ABCD is a trapezium.
AB || CD and AD = BC (Data)
Now, AD || CD and AB || DC.
∴ ADCE is a parallelogram.
∴ AD = CE (Opposite sides)
AD = BC (Data)
CE = BC
∴ ∠CBE = ∠CEB
Now, ∠DAB + ∠CEB = 180° (Sum of consecutive angles.)
∴ ∠DAB + ∠CEB = 180°
( ∵ ∠CEB = ∠CBE) ……………… (i)
Now, ∠ABC + ∠CDE = 180°
(linear pair) ……………….. (ii)
Comparing (i) and (ii),
∠DAB + ∠CBE = ∠ABC + ∠CBE
∴ ∠DAB = ∠ABC Or ∠A = ∠B.

(ii) AB || CD
∠DAB + ∠ADC =180° ……….. (iii)
∠ABC + ∠BCD = 180° ………… (iv)
Compairing (iii) and (iv),
∠DAB + ∠ADC = ∠ABC + ∠BCD
∴ ∠ADC = ∠BCD (∴∠A = ∠B)
Or ∠D = ∠C.

(iii)

Draw diagonals AC and BD.
In ∆ABC and ∆ABD,
BC = AD (Data)
∠ABC = ∠BAD (proved)
AB is common.
∴ ∆ABC ≅ ∆ABD (SAS Postulate).

(iv) ∆ABC ≅ ∆BAD (Proved)
∴ Diagonal AC = Diagonal BD.

We hope the KSEEB Solutions for Class 9 Maths Chapter 7 Quadrilaterals Ex 7.1 helps you. If you have any query regarding Karnataka Board Class 9 Maths Chapter 7 Quadrilaterals Exercise 7.1, drop a comment below and we will get back to you at the earliest.

Karnataka Board Class 9 Maths Chapter 11 Areas of Parallelograms and Triangles Ex 11.2

Question 1.
In Fig., ABCD is a parallelogram, AE ⊥ DC and CF ⊥ AD. AB = 16 cm., AE = 8 cm, CF = 10cm, find AD.

Solution:
ABCD is a parallelogram.
AE ⊥ DC and CF ⊥ AD.
AB = 16 Cm, AE = 8 cm, CF = 10 cm, then AD = ?
(i) Area of ABCD = base × altitude
= DC × AE ( ∵ AB=DC)
= 16 × 8
= 128 sq. cm.
Whether ABCD is quadrilateral, ADCB is a quadrilateral.
∴ Area of ADCB = 128 sq. cm.
Base = AD = ?
Altitude, CF= 100 cm.
Base × Altitude = Area
AD × CF = 128
AD × 10 = 128
∴ AD = \(\frac{128}{10}\)
∴ AD = 12.8 cm.

Question 2.
If E, F, G and H are respectively the mid-points of the sides of a parallelogram ABCD, show that

ar(EFGH) = \(\frac{1}{2}\) ar(ABCD)
Solution:
Data: E, F, G and H are mid-points of the sides of a parallelogram ABCD,
To Prove: area (EFGH) = \(\frac{1}{2}\) area (ABCD)
Construction: HF is joined.
Proof: Now, AD = BC and AD || BC
∴ 2AH = 2BF
∴ AH = BF and AH || BF
∴ AHFB is a parallelogram.
Similarly, HDCF is a parallelogram.
Now, DEHF and Quadrilateral AHFD are on base HF and in between HF || AB.

Question 3.
P and Q are any two points lying on the sides DC and AD respectively of a parallelogram ABCD. Show that ar (PAB) = ar (BQC).

Solution:
Data: P and Q are any two points lying on the sides DC and AD respectively of a parallelogram ABCD.
To Prove: ar(∆APB) = ar(∆BQC)
Proof: ABCD is a parallelogram.
∴ AB || DC AB = DC
AD || BC AD = BC
Now ∆APB and ABCD are on same base AB and in between AB || DC
∴ Area(∆APB) = \(\frac{1}{2}\)  Area ABCD) ……… (i)
Similarly, ∆BQC and BADC are on the same base BC and in between BC || AD.
∴ Area(∆BQC) = \(\frac{1}{2}\)  Area(ABCD) ………. (ii)
From (i) and (ii),
∴ Area(∆APB) = Area (∆BQC).

Question 4.
In Fig., P is a point in the interior of a parallelogram ABCD, Show that

(i) ar.(ABP)= ar(PCD) = \(\frac{1}{2}\) ar (ABCD)
(ii) ar(APD)+ ar(PBC) = ar(APB) + ar(PCD)
(Hint: Through P, draw a line parallel to AB).
Solution:
Data : P is any point in the interior of a parallelogram ABCD. PA, PB, PC and PD are joined.
To Prove: (i) ar(ABP= ar(PCD)= \(\frac{1}{2}\) ar (ABCD)
(ii) ar(APD) + ar(PBC)= ar(APB) + ar(PCD)

Construction: AB || XY is drawn through P.
Proof: (i) XY || AB || DC
∴ ABYX and XDCY are parallelograms.
∆APB and ABYX are on base AB and in between AB||XY.

Construction: AD || MN is drawn through P.
Proof: ∆ADP and ADMN lie on the base AD and in between AD || MN

Question 5.
In Fig., PQRS and ABRS are parallelograms and X is any point on side BR. Show that

(i) ar.(PQRS) = ar.(ABRS)
(ii) ar.(AXS) + = \(\frac{1}{2}\) ar. (PQRS).
Solution:
Data: PQRS and ABRS are parallelograms and X is any point on side BR.

Question 6.
A farmer was having a field in the form of a parallelogram PQRS. She took any point A on RS and joined it to points P and Q. In how many parts the fields is divided? What are the shapes of these parts? The farmer wants to sow wheat and pulses in equal portions of the field separately. How should she do it?

Solution:
If AP and AQ are joined PQRS is divided into three triangles, ∆SAP, ∆APQ and ∆ARQ.
∆APQ and PQRS are on base PQ and in between PQ || SR.

From (i) and (ii).
∴ ar.(∆APQ) = ar.(∆SAP) + ar.(∆ARQ)
∴ Farmer can use the part of the field to sow wheat,
i.e. Ar. (∆SAP + ∆ARQ) and in the same area, he can use the Area of ∆APQ to sow pulses.

KSEEB Solutions for Class 9 Maths

KSEEB Solutions for Class 9 Maths Chapter 6 Constructions Ex 6.2 are part of KSEEB Solutions for Class 9 Maths. Here we have given Karnataka Board Class 9 Maths Chapter 6 Constructions Exercise 6.2.

Karnataka Board Class 9 Maths Chapter 6 Constructions Ex 6.2

Question 1.
Construct a triangle ABC in which BC = 7 cm, ∠B = 75° and AB + AC = 13 cm,
Solution:
Construction of a triangle ABC in which BC = 7 cm, ∠B = 75° and AB + AC = 13 cm

Steps of Construction :

1. Draw BC = 7 cm line segment.
2. Construct ∠B = 75° and produce it upto Bx.
3. With B as centre draw an arc BD = 13 cm such that it intersects Bx at D.
4. Construct PQ which is the perpendicular bisector of BD.
5. PQ intersects BD at A.
6. Join AC.
7. Now ∆ABC, with measurement AB + AC = 13 cm, and ∠B = 75°, BC = 7 cm is constructed.

Question 2.
Construct a triangle ABC in which BC = 8 cm, ∠B = 45° and AB – AC = 3.5 cm.
Solution:
Constructing a triangle ABC in which BC = 8 cm, ∠B = 45° and AB – AC = 3.5 cm.

Steps of Construction :

1. Draw a line segment BC = 8 cm.
2. Draw ∠B = 45° with the help of a protractor. Bx is joined.
3. Draw an arc BD = 3.5 cm.
4. When perpendicular bisector DC is drawn, that intersects Bx at A. AC is joined.
5. Now, DABC with BC = 8 cm, ∠B = 45° and AB – AC = 3.5 cm. is constructed.

Question 3.
Construct a triangle PQR in which QR = 6 cm, ∠Q = 60° and PR – PQ = 2 cm.
Solution:
Constructing a triangle PQR in which QR = 6 cm, ∠Q = 60° and PR – PQ = 2 cm.

Steps of Construction :

1. Draw a line segment QR = 6 cm.
2. With the help of a protractor draw a line segment Qx and ∠Q = 60°.
3. Produce line segment xQ upto y. mark T such that QT = 2 cm. Join TR.
4. If the perpendicular bisector of TR is drawn, that meets line segment Qx at P. PR is joined.
   Now, ∆PQR is constructed.

Question 4.
Construct a triangle XYZ in which ∠Y = 30°, ∠Z = 90° and XY + YZ + ZX = 11 cm.
Solution:
Constructing a triangle XYZ in which ∠Y = 30°, ∠Z = 90° and XY + YZ + ZX = 11 cm.

Steps of Construction :

1. Draw AB = 11 cm line segment with measurement XY + YZ + ZX.
2. With the help of protractor construct ∠A = 30°, ∠B = 90°.
3. Construct angular bisectors of ∠CAB and ∠DBA which meet at x. Ax, Bx are joined.
4. If perpendicular bisector AX is drawn, it intersects AB at Y and AB at Z.
5. By joining XY and XZ, ∆XYZ is constructed.

Question 5.
Construct a right triangle whose base is 12 cm and the sum of its hypotenuse and the other side is 18 cm.
Solution:
Constructing a right-angled triangle ABC with base, AB =12 cm and the sum of hypotenuse BC and another side AC, AC + BC = 18 cm.

Steps of Construction :

1. Draw a line segment AB = 12 cm.
2. With the help of a protractor construct ∠A = 90° and produce upto AX.
3. Mark ‘D’ on AX such that AD = 18 cm.
4. Join BD.
5. Let the perpendicular bisector BD intersects AD at C. Join BC.
6. Now required triangle ABC is constructed with Base AB = 12cm, Side AC = – 5 cm, Hypotenuse BC = 13 cm and ∠A = 90°.

We hope the KSEEB Solutions for Class 9 Maths Chapter 6 Constructions Ex 6.2 help you. If you have any query regarding Karnataka Board Class 9 Maths Chapter 6 Constructions Exercise 6.2, drop a comment below and we will get back to you at the earliest.