Advertisements
Advertisements
Question
In a right-angled triangle PQR right-angled at Q, QR = x cm, PQ = (x + 7) cm and area = 30 cm2. Find the sides of the triangle.
Solution
Area of right-angled ΔPQR = `(1)/(2) xx "QR" xx "PQ"`
⇒ 30 = `(1)/(2) xx x xx (x + 7)`
⇒ 60 = x2 + 7x
⇒ x2 + 7x - 60 = 0
⇒ x2 + 12x - 5x - 60 = 0
⇒ x(x + 12) -5(x + 12) = 0
⇒ (x + 12)(x - 5) = 0
⇒ x + 12 = 0 x - 5 = 0
⇒ x = -12 or x = 5
But, length of a side cannot be negatice.
Hence,
x = 5cm = QR
x + 7
= 5 + 7
= 12cm
= PQ
In right-angled ΔPQR, by Pythagoras theorem,
PR2 = PQ2 + QR2
= 122 + 52
= 144 + 25
= 169
⇒ PR = 13cm
Hence, PQ = 12cm, QR = 5cm and PR = 13cm.
APPEARS IN
RELATED QUESTIONS
Find the area of an equilateral triangle of side 20 cm.
Find the area of the shaded region in the figure as shown, in which DPQS is an equilateral triangle and ∠PQR = 90°.
In a right-angled triangle ABC, if ∠B = 90°, AB - BC = 2 cm; AC - BC = 4 and its perimeter is 24 cm, find the area of the triangle.
In a right-angled triangle ABC, if ∠B = 90°, AB - BC = 2 cm; AC - BC = 4 and its perimeter is 24 cm, find the area of the triangle.
Find the area of an isosceles triangle whose perimeter is 50cm and the base is 24cm.
Find the area of an isosceles triangle whose perimeter is 72cm and the base is 20cm.
Each equal side of an isosceles triangle is 3cm less than the unequal side. The height of the perpendicular from the vertex to the unequal side is 3cm less than the equal side. If the area of the isosceles triangle is 108cm2, find the perimeter of the triangle.
Each of the equal sides of an isosceles triangle is 4 cm greater than its height. If the base of the triangle is 24 cm; calculate the perimeter and the area of the triangle.
If in an isosceles triangle, each of the base angles is 40°, then the triangle is ______.
In an isosceles triangle, two angles are always ______.
