Advertisements
Advertisements
प्रश्न
Find the length of the perpendicular of a triangle whose base is 5cm and the hypotenuse is 13cm. Also, find its area.
Advertisements
उत्तर
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.
APPEARS IN
संबंधित प्रश्न
ABCD is a rhombus. Prove that AB2 + BC2 + CD2 + DA2= AC2 + BD2
Sides of triangle are given below. Determine it is a right triangle or not? In case of a right triangle, write the length of its hypotenuse. 7 cm, 24 cm, 25 cm
In the following figure, O is a point in the interior of a triangle ABC, OD ⊥ BC, OE ⊥ AC and OF ⊥ AB. Show that
(i) OA2 + OB2 + OC2 − OD2 − OE2 − OF2 = AF2 + BD2 + CE2
(ii) AF2 + BD2 + CE2 = AE2 + CD2 + BF2
Two poles of heights 6 m and 11 m stand on a plane ground. If the distance between the feet of the poles is 12 m, find the distance between their tops.
In an equilateral triangle ABC, D is a point on side BC such that BD = `1/3BC` . Prove that 9 AD2 = 7 AB2
In ∆ABC, ∠BAC = 90°, seg BL and seg CM are medians of ∆ABC. Then prove that:
4(BL2 + CM2) = 5 BC2
In ∆ABC, seg AD ⊥ seg BC, DB = 3CD.
Prove that: 2AB2 = 2AC2 + BC2
In right angle ΔABC, if ∠B = 90°, AB = 6, BC = 8, then find AC.
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.
In a quadrilateral ABCD, ∠B = 90° and ∠D = 90°.
Prove that: 2AC2 - AB2 = BC2 + CD2 + DA2
A ladder, 6.5 m long, rests against a vertical wall. If the foot of the ladder is 2.5 m from the foot of the wall, find up to how much height does the ladder reach?
In the right-angled ∆LMN, ∠M = 90°. If l(LM) = 12 cm and l(LN) = 20 cm, find the length of seg MN.
Find the Pythagorean triplet from among the following set of numbers.
3, 4, 5
In a triangle ABC, AC > AB, D is the midpoint BC, and AE ⊥ BC. Prove that: AB2 + AC2 = 2AD2 + `(1)/(2)"BC"^2`
AD is perpendicular to the side BC of an equilateral ΔABC. Prove that 4AD2 = 3AB2.
To get from point A to point B you must avoid walking through a pond. You must walk 34 m south and 41 m east. To the nearest meter, how many meters would be saved if it were possible to make a way through the pond?
In triangle ABC, line I, is a perpendicular bisector of BC.
If BC = 12 cm, SM = 8 cm, find CS
The perimeters of two similar triangles ABC and PQR are 60 cm and 36 cm respectively. If PQ = 9 cm, then AB equals ______.
In an isosceles triangle PQR, the length of equal sides PQ and PR is 13 cm and base QR is 10 cm. Find the length of perpendicular bisector drawn from vertex P to side QR.
Two angles are said to be ______, if they have equal measures.
