Advertisements
Advertisements
Question
Prove that, in a right-angled triangle, the square of the hypotenuse is equal to the sum of the square of remaining two sides.
Advertisements
Solution 1
Draw perpendicular BD from the vertex B to the side AC. A – D
In right-angled ΔABC
seg BD ⊥ hypotenuse AC.
∴ By the similarity in right-angled triangles
ΔABC ~ ΔADB ~ ΔBDC
Now, ΔABC ~ ΔADB
∴ `"AB"/"AD" = "AC"/"AB"` ...(c.s.s.t)
∴ AB2 = AC × AD ...(1)
Also, ΔABC ~ ΔBDC
∴ `"BC"/"DC" = "AC"/"BC"` ...(c.s.s.t)
∴ BC2 = AC × DC ...(2)
From (1) and (2),
AB2 + BC2 = AC × AD + AC × DC
= AC × (AD + DC)
= AC × AC ...(A – D – C)
∴ AB2 + BC2 = AC2
i.e., AC2 = AB2 + BC2
Solution 2
Given: In ΔPQR, ∠PQR = 90°.
To prove: PR2 = PQ2 + QR2.
Construction:
Draw seg QS ⊥ side PR such that P–S–R.
Proof: In ΔPQR,
∠PQR = 90° ...(Given)
Seg QS ⊥ hypotenuse PR ...(Construction)
∴ ΔPSQ ∼ ΔQSR ∼ ΔPQR ...(Similarity of right-angled triangles) ...(1)
ΔPSQ ∼ ΔPQR ...[From (1)]
∴ `"PS"/"PQ" = "PQ"/"PR"` ...(Corresponding sides of similar triangles are in proportion)
∴ PQ2 = PS × PR ...(2)
ΔQSR ∼ ΔPQR ...[From (1)]
∴ `"SR"/"QR" = "QR"/"PR"` ...(Corresponding sides of similar triangles are in proportion)
∴ QR2 = SR × PR ...(3)
Adding (2) and (3), we get
PQ2 + QR2 = PS × PR + SR × PR
∴ PQ2 + QR2 = PR(PS + SR)
∴ PQ2 + QR2 = PR × PR ...(P–S–R)
∴ PQ2 + QR2 = PR2 OR PR2 = PQ2 + QR2.
RELATED QUESTIONS
Side of a triangle is given, determine it is a right triangle.
`(2a – 1) cm, 2\sqrt { 2a } cm, and (2a + 1) cm`
PQR is a triangle right angled at P and M is a point on QR such that PM ⊥ QR. Show that PM2 = QM . MR
ABC is an isosceles triangle with AC = BC. If AB2 = 2AC2, prove that ABC is a right triangle.
A guy wire attached to a vertical pole of height 18 m is 24 m long 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 taut?
D and E are points on the sides CA and CB respectively of a triangle ABC right angled at C. Prove that AE2 + BD2 = AB2 + DE2
Identify, with reason, if the following is a Pythagorean triplet.
(4, 9, 12)
Find the length of the hypotenuse of a right angled triangle if remaining sides are 9 cm and 12 cm.
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 triangle ABC, given below, AB = 8 cm, BC = 6 cm and AC = 3 cm. Calculate the length of OC.
In a rectangle ABCD,
prove that: AC2 + BD2 = AB2 + BC2 + CD2 + DA2.
If the angles of a triangle are 30°, 60°, and 90°, then shown that the side opposite to 30° is half of the hypotenuse, and the side opposite to 60° is `sqrt(3)/2` times of the hypotenuse.
Prove that (1 + cot A - cosec A ) (1 + tan A + sec A) = 2
In ∆ ABC, AD ⊥ BC.
Prove that AC2 = AB2 +BC2 − 2BC x BD
Triangle XYZ is right-angled at vertex Z. Calculate the length of YZ, if XY = 13 cm and XZ = 12 cm.
Find the length of the hypotenuse of a triangle whose other two sides are 24cm and 7cm.
A ladder 25m long reaches a window of a building 20m above the ground. Determine the distance of the foot of the ladder from the building.
Two trains leave a railway station at the same time. The first train travels due west and the second train due north. The first train travels at a speed of `(20 "km")/"hr"` and the second train travels at `(30 "km")/"hr"`. After 2 hours, what is the distance between them?
Find the length of the support cable required to support the tower with the floor
A flag pole 18 m high casts a shadow 9.6 m long. Find the distance of the top of the pole from the far end of the shadow.
Two circles having same circumference are congruent.
