English

Find the Length of the Perpendicular of a Triangle Whose Base is 5cm and the Hypotenuse is 13cm. Also, Find Its Area.

Advertisements
Advertisements

Question

Find the length of the perpendicular of a triangle whose base is 5cm and the hypotenuse is 13cm. Also, find its area.

Sum
Advertisements

Solution

Base = 5cm, Hypotenuse = 13cm
By Pythagoras theorem,
(perpendicular)2 = (13cm)2 - (5cm)2
(perpendicular)2 = 169cm2 - 25cm2
(perpendicular)2 = 144cm2
(perpendicular)2 = (12cm)2
∴ Perpendicular = 12cm
Area of the triangle 
= 13cm2 x (Base x Perpendicular)

= `(1)/(2) xx 5"cm" xx 12"cm"`
= 30cm2.

shaalaa.com
  Is there an error in this question or solution?
Chapter 17: Pythagoras Theorem - Exercise 17.1

APPEARS IN

Frank Mathematics [English] Class 9 ICSE
Chapter 17 Pythagoras Theorem
Exercise 17.1 | Q 1

RELATED QUESTIONS

The points A(4, 7), B(p, 3) and C(7, 3) are the vertices of a  right traingle ,right-angled at B. Find the values of p.


In Figure, ABD is a triangle right angled at A and AC ⊥ BD. Show that AB2 = BC × BD


In Figure ABD is a triangle right angled at A and AC ⊥ BD. Show that AC2 = BC × DC


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


ABC is an isosceles triangle with AC = BC. If AB2 = 2AC2, prove that ABC is a right triangle.


Which of the following can be the sides of a right triangle?

2.5 cm, 6.5 cm, 6 cm

In the case of right-angled triangles, identify the right angles.


Identify, with reason, if the following is a Pythagorean triplet.
(11, 60, 61)


Find the length diagonal of a rectangle whose length is 35 cm and breadth is 12 cm.


Walls of two buildings on either side of a street are parallel to each other. A ladder 5.8 m long is placed on the street such that its top just reaches the window of a building at the height of 4 m. On turning the ladder over to the other side of the street, its top touches the window of the other building at a height 4.2 m. Find the width of the street.


In the given figure, point T is in the interior of rectangle PQRS, Prove that, TS+ TQ= TP+ TR(As shown in the figure, draw seg AB || side SR and A-T-B)


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.


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


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


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


Triangle ABC is right-angled at vertex A. Calculate the length of BC, if AB = 18 cm and AC = 24 cm.


Triangle PQR is right-angled at vertex R. Calculate the length of PR, if: PQ = 34 cm and QR = 33.6 cm.


In a triangle ABC, AC > AB, D is the midpoint BC, and AE ⊥ BC. Prove that: AB2 + AC2 = 2(AD2 + CD2)


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


In a right-angled triangle ABC,ABC = 90°, AC = 10 cm, BC = 6 cm and BC produced to D such CD = 9 cm. Find the length of AD.


If length of sides of a triangle are a, b, c and a2 + b2 = c2, then which type of triangle it is?


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×