Advertisements
Advertisements
प्रश्न
In the given 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.
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
A(4, - 6), B(3,- 2) and C(5, 2) are the vertices of a 8 ABC and AD is its median. Prove that the median AD divides Δ ABC into two triangles of equal areas.
If the coordinates of two points A and B are (3, 4) and (5, – 2) respectively. Find the coordniates of any point P, if PA = PB and Area of ∆PAB = 10
Find the area of the triangle whose vertices are: (–5, –1), (3, –5), (5, 2)
Find equation of line joining (1, 2) and (3, 6) using the determinant.
Prove that the points A (a,0), B( 0,b) and C (1,1) are collinear, if `( 1/a+1/b) =1`.
In ∆PQR, PR = 8 cm, QR = 4 cm and PL = 5 cm. 
Find:
(i) the area of the ∆PQR
(ii) QM.
The points (1,1), (-2, 7) and (3, -3) are ______.
Find the values of k if the points A(k + 1, 2k), B(3k, 2k + 3) and C(5k – 1, 5k) are collinear.
A rectangular plot is given for constructing a house, having a measurement of 40 m long and 15 m in the front. According to the laws, a minimum of 3 m, wide space should be left in the front and back each and 2 m wide space on each of other sides. Find the largest area where house can be constructed.
Find the missing value:
| Base | Height | Area of Triangle |
| ______ | 31.4 mm | 1256 mm2 |
