## ML Aggarwal Class 9 Solutions for ICSE Maths Chapter 12 Pythagoras Theorem

**Question 1.**

**Lengths of sides of triangles are given below. Determine which of them are right triangles. In case of a right triangle, write the length of its hypotenuse:**

**(i) 3 cm, 8 cm, 6 cm**

**(ii) 13 cm, .12 cm, 5 cm**

**(iii) 1.4 cm, 4.8 cm, 5 cm**

**Solution:**

**Question 2.**

**Foot of a 10 m long ladder leaning against a vertical well is 6 m away from the base of the wail. Find the height of the point on the wall where the top of the ladder reaches.**

**Solution:**

**Question 3.**

**A guy attached a wire 24 m long to a vertical pole of height 18 m and has a stake attached to the other end. How far from the base of the pole should the stake be driven so that the wire will be taught?**

**Solution:**

**Question 4.**

**Two poles of heights 6 m and 11 m stand on a plane ground. If the distance between their feet is 12 m, find the distance between their tops.**

**Solution:**

**Question 5.**

**In a right-angled triangle, if hypotenuse is 20 cm and the ratio of the other two sides is 4:3, find the sides.**

**Solution:**

**Question 6.**

**If the sides of a triangle are in the ratio 3:4:5, prove that it is right-angled triangle.**

**Solution:**

**Question 7.**

**For going to a city B from city A, there is route via city C such that AC ⊥ CB, AC = 2x km and CB=2(x+ 7) km. It is proposed to construct a 26 km highway which directly connects the two cities A and B. Find how much distance will be saved in reaching city B from city A after the construction of highway.**

**Solution:**

**Question 8.**

**The hypotenuse of right triangle is 6m more than twice the shortest side. If the third side is 2m less than the hypotenuse, find the sides of the triangle.**

**Solution:**

**Question 9.**

**ABC is an isosceles triangle right angled at C. Prove that AB² = 2AC².**

**Solution:**

**Question 10.**

**In a triangle ABC, AD is perpendicular to BC. Prove that AB² + CD² = AC² + BD².**

**Solution:**

**Question 11.**

**In ∆PQR, PD ⊥ QR, such that D lies on QR. If PQ = a, PR = b, QD = c and DR = d, prove that (a + b) (a – b) = (c + d) (c – d).**

**Solution:**

**Question 12.**

**ABC is an isosceles triangle with AB = AC = 12 cm and BC = 8 cm. Find the altitude on BC and Hence, calculate its area.**

**Solution:**

**Question 13.**

**Find the area and the perimeter of a square whose diagonal is 10 cm long.**

**Solution:**

**Question 14.**

**(a) In fig. (i) given below, ABCD is a quadrilateral in which AD = 13 cm, DC = 12 cm, BC = 3 cm, ∠ ABD = ∠BCD = 90°. Calculate the length of AB.**

**(b) In fig. (ii) given below, ABCD is a quadrilateral in which AB = AD, ∠A = 90° =∠C, BC = 8 cm and CD = 6 cm. Find AB and calculate the area of ∆ ABD.**

**Solution:**

**Question 15.**

**(a) In figure (i) given below, AB = 12 cm, AC = 13 cm, CE = 10 cm and DE = 6 cm.Calculate the length of BD.**

**(b) In figure (ii) given below, ∠PSR = 90°, PQ = 10 cm, QS = 6 cm and RQ = 9 cm. Calculate the length of PR.**

**(c) In figure (iii) given below, ∠ D = 90°, AB = 16 cm, BC = 12 cm and CA = 6 cm. Find CD.**

**Solution:**

**Question 16.**

**(a) In figure (i) given below, BC = 5 cm,**

**∠B =90°, AB = 5AE, CD = 2AE and AC = ED. Calculate the lengths of EA, CD, AB and AC.**

**(b) In the figure (ii) given below, ABC is a right triangle right angled at C. If D is mid-point of BC, prove that AB2 = 4AD² – 3AC².**

**Solution:**

**Question 17.**

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

**Solution:**

**Question 18.**

**In a rhombus, If diagonals are 30 cm and 40 cm, find its perimeter.**

**Solution:**

**Question 19.**

**(a) In figure (i) given below, AB || DC, BC = AD = 13 cm. AB = 22 cm and DC = 12cm. Calculate the height of the trapezium ABCD.**

**(b) In figure (ii) given below, AB || DC, ∠ A = 90°, DC = 7 cm, AB = 17 cm and AC = 25 cm. Calculate BC.**

**(c) In figure (iii) given below, ABCD is a square of side 7 cm. if**

**AE = FC = CG = HA = 3 cm,**

**(i) prove that EFGH is a rectangle.**

**(ii) find the area and perimeter of EFGH.**

**Solution:**

**Question 20.**

**AD is perpendicular to the side BC of an equilateral Δ ABC. Prove that 4AD² = 3AB².**

**Solution:**

**Question 21.**

**In figure (i) given below, D and E are mid-points of the sides BC and CA respectively of a ΔABC, right angled at C.**

**Solution:**

**Question 22.**

**If AD, BE and CF are medians of ΕABC, prove that 3(AB² + BC² + CA²) = 4(AD² + BE² + CF²).**

**Solution:**

**Question 23.**

**(a) In fig. (i) given below, the diagonals AC and BD of a quadrilateral ABCD intersect at O, at right angles. Prove that**

**AB² + CD² = AD² + BC².**

**(b) In figure (ii) given below, OD⊥BC, OE ⊥CA and OF ⊥ AB. Prove that :**

**(i) OA² + OB² + OC² = AF² + BD² + CE² + OD² + OE² + OF².**

**(ii) OAF² + BD² + CE² = FB² + DC² + EA².**

**Solution:**

**Question 24.**

**In a quadrilateral, ABCD∠B = 90° = ∠D. Prove that 2 AC² – BC2 = AB² + AD² + DC².**

**Solution:**

**Question 25.**

**In a ∆ ABC, ∠ A = 90°, CA = AB and D is a point on AB produced. Prove that :**

**DC² – BD² = 2AB. AD.**

**Solution:**

**Question 26.**

**In an isosceles triangle ABC, AB = AC and D is a point on BC produced. Prove that AD² = AC² + BD.CD.**

**Solution:**

**Question P.Q.**

**(a) In figure (i) given below, PQR is a right angled triangle, right angled at Q. XY is parallel to QR. PQ = 6 cm, PY = 4 cm and PX : OX = 1:2. Calculate the length of PR and QR.**

**(b) In figure (ii) given below, ABC is a right angled triangle, right angled at B.DE || BC.AB = 12 cm, AE = 5 cm and AD : DB = 1: 2. Calculate the perimeter of A ABC.**

**(c)In figure (iii) given below. ABCD is a rectangle, AB = 12 cm, BC – 8 cm and E is a point on BC such that CE = 5 cm. DE when produced meets AB produced at F.**

**(i) Calculate the length DE.**

**(ii) Prove that ∆ DEC ~ AEBF and Hence, compute EF and BF.**

**Solution:**

**Multiple Choice Questions**

**Choose the correct answer from the given four options (1 to 7):**

**Question 1.**

**In a ∆ABC, if AB = 6√3 cm, BC = 6 cm and AC = 12 cm, then ∠B is**

**(a) 120°**

**(b) 90°**

**(c) 60°**

**(d) 45°**

**Solution:**

**Question 2.**

**If the sides of a rectangular plot are 15 m and 8 m, then the length of its diagonal is**

**(a) 17 m**

**(b) 23 m**

**(c) 21 m**

**(d) 17 cm**

**Solution:**

**Question 3.**

**The lengths of the diagonals of a rhombus are 16 cm and 12 cm. The length of the side of the rhombus is**

**(a) 9 cm**

**(b) 10 cm**

**(c) 8 cm**

**(d) 20 cm**

**Solution:**

**Question 4.**

**If a side of a rhombus is 10 cm and one of the diagonals is 16 cm, then the length of the other diagonals is**

**(a) 6 cm**

**(b) 12 cm**

**(c) 20 cm**

**(d) 12 cm**

**Solution:**

**Question 5.**

**If a ladder 10 m long reaches a window 8 m above the ground, then the distance of the foot of the ladder from the base of the wall is**

**(a) 18 m**

**(b) 8 m**

**(c) 6 m**

**(d) 4 m**

**Solution:**

**Question 6.**

**A girl walks 200 m towards East and then she walks ISO m towards North. The distance of the girl from the starting point is**

**(a) 350 m**

**(b) 250 m**

**(c) 300 m**

**(d) 225 m**

**Solution:**

**Question 7.**

**A ladder reaches a window 12 m above the ground on one side of the street. Keeping its foot at the same point, the ladder is turned to the other side of the street to reach a window 9 m high. If the length of the ladder is 15 m, then the width of the street is**

**(a) 30 m**

**(b) 24 m**

**(c) 21 m**

**(d) 18 m**

**Solution:**

**Chapter Test**

**Question 1.**

**(a) In fig. (i) given below, AD ⊥ BC, AB = 25 cm, AC = 17 cm and AD = 15 cm. Find the length of BC.**

**(b) In figure (ii) given below, ∠BAC = 90°, ∠ADC = 90°, AD = 6 cm, CD = 8 cm and BC = 26 cm. Find :**

**(i) AC (ii) AB (iii) area of the shaded region.**

**(c) In figure (iii) given below, triangle ABC is right angled at B. Given that AB = 9 cm, AC = 15 cm and D, E are mid-points of the sides AB and AC respectively, calculate**

**(i) the length of BC (ii) the area of ∆ ADE.**

**Solution:**

**Question 2.**

**If in ∆ ABC, AB > AC and ADI BC, prove that AB² – AC² = BD² – CD².**

**Solution:**

**Question 3.**

**In a right angled triangle ABC, right angled at C, P and Q are the points on the sides CA and CB respectively which divide these sides in the ratio 2:1. Prove that**

**(i) 9AQ² = 9AC² + 4BC²**

**(ii) 9BP² = 9BC² + 4AC²**

**(iii) 9(AQ² + BP²) = 13AB².**

**Solution:**

A right angled ∆ ABC in which ∠ C

**Question 4.**

**In the given figure, ∆PQR is right angled at Q and points S and T trisect side QR. Prove that 8PT² – 3PR² + 5PS².**

**Solution:**

**Question 5.**

**In a quadrilateral ABCD, ∠B = 90°. If AD² = AB² + BC² + CD², prove that ∠ACD = 90°.**

**Solution:**

**Question 6.**

**In the given figure, find the length of AD in terms of b and c.**

**Solution:**

**Question 7.**

**ABCD is a square, F is mid-point of AB and BE is one-third of BC. If area of ∆FBE is 108 cm², find the length of AC.**

**Solution:**

**Question 8.**

**In a triangle ABC, AB = AC and D is a point on side AC such that BC²**** = AC x CD, Prove that BD = BC.**

**Solution:**

मिहिर मिथुन वनमाली says

Thanks for help

Sneha Senapati says

Thank u so much… I am having exam and so many doubts.