RBSE Solutions for Class 9 Maths Chapter 10 Area of Triangles and Quadrilaterals Ex 10.3 is part of RBSE Solutions for Class 9 Maths. Here we have given Rajasthan Board RBSE Class 9 Maths Solutions Chapter 10 Area of Triangles and Quadrilaterals Ex 10.3.

Board |
RBSE |

Class |
Class 9 |

Subject |
Maths |

Chapter |
Chapter 10 |

Chapter Name |
Area of Triangles and Quadrilaterals |

Exercise |
Ex 10.3 |

Number of Questions Solved |
15 |

Category |
RBSE Solutions |

## Rajasthan Board RBSE Class 9 Maths Solutions Chapter 10 Area of Triangles and Quadrilaterals Ex 10.3

Question 1.

ABCD is a parallelogram and X is the mid-point of AB. If ar (AXCD) = 24 cm², then ar (∆ABC) = 24 cm². Is this statement true or false. Give reason for your answer.

Solution.

ABCD is a parallelogram and X is the mid-point of AB. Draw the diagonal AC.

We know that diagonal of (RBSESolutions.com) a parallelogram divides it into two triangles of equal areas.

So, ar (||gm ABCD) = ar (∆ADC) x 2

= 2 x 24 = 48 cm²

Since, CX is the median of ∆ABC

∴ ar (∆BCX) = \(\frac { 1 }{ 2 }\) ar (∆ABC)

= \(\frac { 1 }{ 2 }\) x 24 = 12 cm²

Now,

ar (∆XCD) = ar(||gm ABCD) – ar (∆BXC)

= 48 – 12 = 36 cm²

Hence, the given statement is false.

Question 2.

PQRS is a rectangle inscribed in a quadrant (RBSESolutions.com) of a circle of radius 13 cm. A is any point on PQ. If PS = 5 cm, then ar (∆PAS) = 30 cm².

Solution.

In ∆PQR,

PR² = PQ² + RQ²

(by Pythagoras theorem)

=> PQ² = PR² – RQ² = (13)² – (5)²

(given PS = 5 cm, PR = 13 cm)

=> PQ = √169-25 = √144 = 12 cm

Now, area of rectangle PQRS

= 12 x 5 = 60 cm²

Area of ∆PSQ = \(\frac { 1 }{ 2 }\) x 60 = 30 cm²

(Since, diagonal of a rectangle bisects it into two triangles of equal areas)

Now, A is any point on PQ. So, area of ∆APS depends on position of A. If A is at point Q, then area will be 30 cm².

∴ ar (∆PAS) ≤ 30 cm².

Hence proved.

Question 3.

PQRS is a parallelogram, whose area is 180 cm² and A is any point (RBSESolutions.com) on the diagonal QS. Then, the area of ∆ASR = 90 cm². This statement is true or false. Why?

Solution.

This statement is False.

Given, area of parallelogram PQRS is 180 cm² and QS is its diagonal which divides it into two triangles of equal areas.

∴ Area of ∆PQS = 90 cm²

Now, A is any point on QS.

∴ Area of ∆ASR < Area of ∆SRQ.

Question 4.

∆ABC and ∆BDE are two equilateral triangles such (RBSESolutions.com) that D is the mid-point of BC, then

ar (∆BDE) = \(\frac { 1 }{ 4 }\) ar (∆ABC).

Solution.

Given: In a ∆ABC, AD is the median and E is mid-point of AD.

To prove: ar (∆BDE) = \(\frac { 1 }{ 4 }\) ar (∆ABC).

Proof: ∵ AD is the median of ∆ABC

∴ ar (∆ABD) = \(\frac { 1 }{ 2 }\) ar (∆ABC) …(i)

Reason: Median divides the (RBSESolutions.com) triangle into two triangles equal in area.

Also in ∆BAD, BE is median.

∴ ar (∆BED) = \(\frac { 1 }{ 2 }\) ar (∆ABD) …(ii)

Using (i) and (ii), we get

area (∆BED) = \(\frac { 1 }{ 2 }\) X \(\frac { 1 }{ 2 }\) ar (∆ABC)

Yes, the given statement is true.

Question 5.

In figure, ABCD and EFGD are (RBSESolutions.com) two parallelograms and G is the mid-point of CD, then

ar (∆DPC) = \(\frac { 1 }{ 2 }\) ar (EFGD)

Solution.

Given: ABCD and EFGD are ||gm and G is mid-point of CD.

To prove: ar (∆DPC) = \(\frac { 1 }{ 2 }\) ar (EFGD)

Proof: ar (∆DPC) = \(\frac { 1 }{ 2 }\) ar (||gm ABCD)

(∵ ∆PDC and parallelogram ABCD are on the the same base DC and between the same parallels AB and CD)

∵ G is mid-point of DC.

∴DG = \(\frac { 1 }{ 2 }\)DC

∴ar (EFGD) = \(\frac { 1 }{ 2 }\) ar (ABCD) …(ii)

=> ar (∆DPC) = ar (EFGD)

The given statement is false.

Question 6.

In a trapezium ABCD, AB || DC and L is the (RBSESolutions.com) mid-point of BC. Through L a line PQ is drawn parallel to AD which meets DC produced at Q.

Prove that ar (ABCD) = ar (APQD).

Solution.

In ∆LPB and ∆CLQ

∠1 = ∠2

(Vertically opposite angles)

∠3 = ∠4 (alternate angles)

and BL = LC

(as L is mid-point of BC)

∴∆LPB ≅ ∆LQC

(by ASA congruency rule)

⇒ ar (∆LPB) + ar (∆DCLP)

⇒ ar (∆CLQ) + ar (∆DCLP)

⇒ ar (∆BCD) = ar (∆PQD)

Hence proved.

Question 7.

If the mid-point of the sides of (RBSESolutions.com) a quadrilateral are joined in order, then prove that the area of the parallelogram so formed will be half of the area of the given quadrilateral (see figure).

[Hint: Join BD and draw perpendicular from A on BD]

Solution.

Given: ABCD is (RBSESolutions.com) a quadrilateral and P, F, R and S are mid-points of the sides BC, CD, AD and AB respectively and PFRS is a parallelogram.

To prove:

ar (||gm PFRS) = \(\frac { 1 }{ 2 }\)ar (quad. ABCD)

Construction: Join BD and BR.

Proof: Median BR divides ∆BDA into two triangles of equal areas.

ar (∆BRA) = \(\frac { 1 }{ 2 }\) ar (∆BDA) ..(i)

Similarly, median RS divides ∆BRA into (RBSESolutions.com) two triangles of equal areas.

ar (∆ASR) = \(\frac { 1 }{ 2 }\) ar (∆BRA) …(ii)

From equation (i) and (ii), we get

ar (∆ASR) = \(\frac { 1 }{ 4 }\) ar (∆BDA) …(iii)

Similarly,

ar (∆CFP) = \(\frac { 1 }{ 4 }\) ar (∆BCD) …(iv)

On adding equations (iii) and (iv), we get

ar (∆ASR) + ar (∆CFP) = \(\frac { 1 }{ 4 }\) [ar (∆BDA) + ar (∆BCD)

ar (∆ASR) + ar (∆CFP) = \(\frac { 1 }{ 4 }\) ar (quadrilateral ABCD) …(v)

Similarly, ar (∆DRF) + ar (∆BSP) = \(\frac { 1 }{ 2 }\) ar (quadrilateral ABCD) …(vi)

On adding (v) and (vi), we get

ar (∆ASR) + ar (∆CFP) + ar (∆DRF) + ar (ABSP)

= \(\frac { 1 }{ 2 }\) ar (quadrilateral ABCD) …..(vii)

But

ar (∆ASR) + ar (∆CFP) + ar (∆DRF) + ar (∆BSP) + ar (||gm PFRS)

= ar (quadrilateral ABCD) …(viii)

On subtracting equation (vii) from equation (viii), we get

ar (||gm PFRS) = \(\frac { 1 }{ 2 }\) ar

(quadrilateral ABCD)

Hence proved.

Question 8.

A man walks 10 m towards (RBSESolutions.com) East and then he walks 30 m towards North. Find his distance from the starting point.

Solution.

Question 9.

A ladder is placed in such a way that its foot is (RBSESolutions.com) at a distance of 7 m from the wall. If its other end reaches a window, a height of 24 m, then find the length of the ladder.

Solution.

In ∆ABC,

AC² = AB² + BC²

(Using Baudhayan theorem)

AC² = (24)² + (7)²

= 576 + 49 = 625

=> AC = √625 = 25 m

Hence, length of the ladder = 25 m.

Question 10.

Two poles stand vertically on (RBSESolutions.com) a level ground. Their height are 7 m and 12 m respectively. If the distance between their feets is 12 m. Find the distance between their tops.

Solution.

Here AB = 7 m and CD = 12 m

CP = (12 – 7) m = 5 m

∴In right angled ∆APC

AC² = AP² + PC²

=> AC² = (12 m)² + (5 m)²

=> AC² = 144 m² + 25 m²

=> AC = √169 m

=> AC = 13 m

Hence, distance between their tops is 13 m.

Question 11.

Find the length of the altitude and area of (RBSESolutions.com) an equilateral triangle of side ‘a’

Solution.

∵ ABC is an equilateral triangle.

∴ AB = BC = CA = a

Question 12.

Find the length of the diagonal of (RBSESolutions.com) a square whose each side is of 4 m.

Solution.

Length of diagonal of a square = √2a

= √2×4

= 4√2 m

Question 13.

In an equilateral triangle ABC, AD is perpendicular to BC, prove that 3AB² = 4AD².

Solution.

∵ AD ⊥ BC(given)

Hence proved.

Question 14.

O is a point inside (RBSESolutions.com) the rectangle ABCD. Prove that OB² + OD² = OA² + OC².

Solution.

Given: O is a point inside the rectangle ABCD.

To prove: OB² + OD² = OA² + OC².

Construction: Join O, with the vertices A, B, C and D of rectangle ABCD. Now draw (RBSESolutions.com) a parallel line LM through O which meets AB and DC at M and L respectively.

Proof: In ∆OMB

OB² = OM² + MB² = OM² + CL²

(by Baudhayan theorem)

Since, ABCD is a rectangle and ML ⊥ AB

MB = CL and in ∆ODL

OD² = OL² + DL²

= OL² + AM² (∵ DL = AM)

∴ OB² + OD² = OM² + CL² + OL² + AM²

= (OM² + AM²) + (CL² + OL²)

= OA² + OC²

Hence, OB² + OD² = OA² + OC²

Hence proved.

Question 15.

In an obtuse angled triangle ABC, ∠C is an obtuse angle and AD ⊥ BC which (RBSESolutions.com) meets BC produced at D. Then prove that AB² = AC² + BC² + 2BC • CD.

Solution.

In right angled ∆ADB

Using Baudhayan theorem

AB² = AD² + DB²

or AB² = AD² + (DC + CB)²

or AB² = AD² + DC² + BC² + 2BC • DC

or AB² = (AD² + DC²) + BC² + 2BC • CD

=> AB² = AC² + BC² + 2BC • CD

(In ∆ADC, AC² = AD² + DC²)

We hope the given RBSE Solutions for Class 9 Maths Chapter 10 Area of Triangles and Quadrilaterals Ex 10.3 will help you. If you have any query regarding RBSE Rajasthan Board Solutions for Class 9 Maths Chapter 10 Area of Triangles and Quadrilaterals Ex 10.3, drop a comment below and we will get back to you at the earliest.