Advertisements
Advertisements
प्रश्न
In the following figure, area of ΔPQR is 20 cm2 and area of ΔPQS is 44 cm2. Find the length RS, if PQ is perpendicular to QS and QR is 5 cm.
Advertisements
उत्तर
Given, area of ΔPQR = 20 cm2 and area of ΔPQS = 44 cm2
We know that,
Area of triangle = `1/2` × Base × Height
∴ Area of ΔPQR = `1/2` × PQ × QR ...[∵ PQ ⊥ QR]
⇒ 20 = `1/2` × PQ × 5
⇒ `(20 xx 2)/5` = PQ ...[∵ QR = 5 cm, given]
⇒ PQ = 8 cm
∴ Area of ΔPQS = `1/2` × PQ × QS
⇒ 44 = `1/2` × 8 × QS
⇒ QS = `(44 xx 2)/8` ...[∵ PQ = 8 cm]
⇒ QS = 11 cm
Now, RS = QS – QR = 11 – 5 = 6 cm
APPEARS IN
संबंधित प्रश्न
Determine the ratio in which the line 2x + y – 4 = 0 divides the line segment joining the points A(2, – 2) and B(3, 7).
Find the area of a triangle whose vertices are
(6,3), (-3,5) and (4,2)
Prove that (2, -2) (-2, 1) and (5, 2) are the vertices of a right-angled triangle. Find the area of the triangle and the length of the hypotenuse.
Find the area of ΔABC whose vertices are:
A(-5,7) , B (-4,-5) and C (4,5)
Find the area of ΔABC whose vertices are:
A(10,-6) , B (2,5) and C(-1,-3)
If the points P(-3, 9), Q(a, b) and R(4, -5) are collinear and a+b=1, find the value of a and b.
Using determinants, find the values of k, if the area of triangle with vertices (–2, 0), (0, 4) and (0, k) is 4 square units.
In a triangle ABC, if `|(1, 1, 1),(1 + sin"A", 1 + sin"B", 1 + sin"C"),(sin"A" + sin^2"A", sin"B" + sin^2"B", sin"C" + sin^2"C")|` = 0, then prove that ∆ABC is an isoceles triangle.
Find the area of the triangle whose vertices are (-2, 6), (3, -6), and (1, 5).
Find the values of k if the points A(k + 1, 2k), B(3k, 2k + 3) and C(5k – 1, 5k) are collinear.
