Selina solutions concise maths class 9 icse Solutions for Chapter 13 chapter 13 pythagoras theorem

Exercises in Chapter 13 chapter 13 pythagoras theorem Grade 9

Exercise 13(A)

Exercise 13(B)

Questions in Exercise 13(A)

A ladder 13 m long rests against a vertical wall. If the foot of the ladder is 5m from the foot

of the wall, find the distance of the other end of the ladder from the ground.

A man goes 40m due north then 50 m due west. Find his distance from the starting point.

In the figure: ∠PSQ = 900, PQ= 10 cm, QS = 6cm and RQ = 9 cm. Calculate the length of P.

The given figure shows a quadrilateral ABCD in which AD = 13cm, DC = 12 cm, BC = 3

cm and ∠ABD = ∠BCD = 900. Calculate the length of AB.

AD is drawn perpendicular to base BC of an equilateral triangle ABC. Given BC = 10 cm,

find the length of AD, correct to 1 place of decimal.

In triangle ABC, given below, AB = 8cm, BC = 6cm and Ac = 3cm. Calculate the length of OC.

In triangle ABC,

AB = AC = x; BC = 10cm and the area of the triangle is 60 cm2. Find x.

If the sides of a triangle are in the ratio 1: √𝟐: 1, show that it is a right angles triangle.

A parallelogram ABCD has P the mid-point of DC and Q a point of AC such that CQ = $\frac{1}{4} \mathrm{AC} . \text { PQ produced meets } \mathrm{BC} \text { at } \mathrm{R} .$

Prove that:

(i) R is the mid-point of BC,

$\text { (ii) } \quad \mathrm{PR}=\frac{1}{2} \mathrm{DB}$

Two poles of heights 6 m and 11 m stand vertically on a plane ground. If the distance

between their feet is 12m; find the distance between their tips

In the given figure, AB||CD, AB = 7cm, BD = 25cm and CD = 17 cm; find the length of side

BC.

In the given figure, ∠B=900, XY||BC, AB=12cm, AY = 8cm and AX : XB = 1: 2 = AY : YC.

Find the lengths of AC and BC.

In ΔABC, ∠B = 900. Find the sides of the triangle, if:

(i) AB = (x-3) cm, BC = (x+4) cm and AC = (x+6)cm

(ii) AB = x cm, BC = (4x+4)cm and AC = (4x+5) cm

Questions in Exercise 13(B)

In the figure, given below, AD⊥BC. Prove that: c2=a2+b2-2ax.

An equilateral triangle ABC, AD⊥BC and BC = x cm. Find, in terms of x, the length of AD.

ABC is a triangle, right angles at B. M is a point on BC. Prove that:

$\mathrm{AM}^{2}+\mathrm{BC}^{2}=\mathrm{AC}^{2}+\mathrm{BM}^{2}$

M and N are the mid-points of the sides QR and PQ respectively of a triangle PQR, right

angled at Q. Prove that:

$\text { (i) } \quad \mathrm{PM}^{2}+\mathrm{RN}^{2}=5 \mathrm{MN}^{2}$

$\text { (ii) } 4 \mathrm{PM}^{2}=4 \mathrm{PQ}^{2}+\mathrm{QR}^{2}$

$\text { (iii) } \quad 4 \mathrm{RN}^{2}=\mathrm{PQ}^{2}+4 \mathrm{QR}^{2}$

$\text { (iv) } 4\left(\mathrm{PM}^{2}+\mathrm{RN}^{2}\right)=5 \mathrm{PR}^{2}$

In triangle ABC, ∠B = 900, and D is the midpoint of BC. Prove that:

$A C^{2}=A D^{2}+3 C D^{2}$

In a rectangle ABCD, prove that:

$\mathbf{A} \mathbf{C}^{2}+\mathbf{B D}^{2}=\mathbf{A} \mathbf{B}^{2}+\mathbf{B} \mathbf{C}^{2}+\mathbf{C D}^{2}+\mathbf{D} \mathbf{A}^{2}$

$\begin{aligned} &\text { In a quadrilateral } A B C D, \angle B=90^{\circ} \text { and } \angle D=90^{\circ} \text { . Prove that: }\\ &2 A C^{2}-A B^{2}=B C^{2}+C D^{2}+D A^{2} \end{aligned}$

$\text { O is any point inside a rectangle } A B C D \text { . Prove that: } O B^{2}+O D^{2}=O C^{2}+O A^{2}$

In the following figure, OP, OQ, and OR are drawn perpendicular to the sides BC, CA and

AB respectively of triangle ABC. Prove that:

$A R^{2}+B P^{2}+C Q^{2}=A Q^{2}+C P^{2}+B R^{2}$

Diagonals of rhombus ABCD intersect each other at point O. Prove that:

$\mathrm{OA}^{2}+\mathrm{OC}^{2}=2 \mathrm{AD}^{2}-\frac{B D^{2}}{2}$

In figure AB = BC and AD is perpendicular to CD. Prove that:

$A C^{2}=2 . B C . D C$

$\begin{aligned} &\text { In an isosceles triangle } A B C ; A B=A C \text { and } D \text { is a point on } B C \text { produced. Prove that: }\\ &\mathrm{AD}^{2}=\mathrm{AC}^{2}+\mathrm{BD} \cdot \mathrm{CD} \end{aligned}$

In an isosceles triangle ABC; angle $A=90^{\circ}, C A=A B \text { and } D \text { is a point on } A B \text { produced. }$

$\begin{aligned} &\text { Prove that: }\\ &\mathrm{DC}^{2}-\mathrm{BD}^{2}=2 \mathrm{AB} \cdot \mathrm{AD} \end{aligned}$

$\begin{aligned} &\text { In triangle } \mathrm{ABC}, \mathrm{AB}=\mathrm{AC} \text { and } \mathrm{BD} \text { is perpendicular to } \mathrm{AC} \text { . Prove that: }\\ &\mathrm{BD}^{2}-\mathrm{CD}^{2}=2 \mathrm{CD} \times \mathrm{AD} \end{aligned}$

In the following figure, AD is perpendicular to BC and D divides BC in the ratio 1:3.