Selina solutions for Concise Mathematics [English] Class 9 ICSE chapter 13 - Pythagoras Theorem [Proof and Simple Applications with Converse] [Latest edition]

A ladder 13 m long rests against a vertical wall. If the foot of the ladder is 5 m from the foot of the wall, find the distance of the other end of the ladder from the ground.

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

In the figure: ∠PSQ = 90^o, PQ = 10 cm, QS = 6 cm and RQ = 9 cm. Calculate the length of PR.

The given figure shows a quadrilateral ABCD in which AD = 13 cm, DC = 12 cm, BC = 3 cm and ∠ABD = ∠BCD = 90^o. 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 = 8 cm, BC = 6 cm and AC = 3 cm. Calculate the length of OC.

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

If the sides of the triangle are in the ratio 1: `sqrt2`: 1, show that is a right-angled triangle.

Two poles of heights 6 m and 11 m stand vertically on a plane ground. If the distance between their feet is 12 m;
find the distance between their tips.

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

In the given figure, ∠B = 90^°, XY || BC, AB = 12 cm, AY = 8cm and AX : XB = 1 : 2 = AY : YC.

Find the lengths of AC and BC.

In ΔABC,  Find the sides of the triangle, if:

1. AB =  ( x - 3 ) cm, BC = ( x + 4 ) cm and AC = ( x + 6 ) cm
2. AB = x cm, BC = ( 4x + 4 ) cm and AC = ( 4x + 5) cm

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

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

ABC is a triangle, right-angled at B. M is a point on BC.

Prove that: AM² + BC² = AC² + BM²

M andN are the mid-points of the sides QR and PQ respectively of a PQR, right-angled at Q.
Prove that:
(i) PM² + RN² = 5 MN²
(ii) 4 PM² = 4 PQ² + QR²
(iii) 4 RN² = PQ² + 4 QR²(iv) 4 (PM² + RN²) = 5 PR²

In triangle ABC, ∠B = 90^o and D is the mid-point of BC.

Prove that: AC² = AD² + 3CD².

In a rectangle ABCD,
prove that: AC² + BD² = AB² + BC² + CD² + DA².

In a quadrilateral ABCD, ∠B = 90° and ∠D = 90°.
Prove that: 2AC² - AB² = BC² + CD² + DA²

O is any point inside a rectangle ABCD.
Prove that: OB² + OD² = OC² + OA².

In the following figure, OP, OQ, and OR are drawn perpendiculars to the sides BC, CA and AB respectively of triangle ABC.

Prove that: AR² + BP² + CQ² = AQ² + CP² + BR²

Diagonals of rhombus ABCD intersect each other at point O.

Prove that: OA² + OC² = 2AD² - `"BD"^2/2`

In figure AB = BC and AD is perpendicular to CD.
Prove that: AC² = 2BC. DC.

In an isosceles triangle ABC; AB = AC and D is the point on BC produced.
Prove that: AD² = AC² + BD.CD.

In triangle ABC, angle A = 90^o, CA = AB and D is the point on AB produced.
Prove that DC² - BD² = 2AB.AD.

In triangle ABC, AB = AC and BD is perpendicular to AC.

Prove that: BD² - CD² = 2CD × AD

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

Prove that : 2AC² = 2AB² + BC²