Advertisements
Advertisements
Question
PQRS is a rectangle inscribed in a quadrant of a circle of radius 13 cm. A is any point on PQ. If PS = 5 cm, then ar (PAS) = 30 cm2.
Options
True
False
Advertisements
Solution
This statement is True.
Explanation:
Given, PS = 5 cm
Radius of circle = SQ = 13 cm
In right-angled ΔSPQ,
SQ2 = PQ2 + PS2 ...[By Pythagoras theorem]
(13)2 = PQ2 + (5)2
⇒ PQ2 = 169 – 25 = 144
⇒ PQ = 12 cm ...[Taking positive square root, because length is always positive]
Now, area of ΔAPS = `1/2` × Base × Height
= `1/2 xx PS xx PQ`
= `1/2 xx 5 xx 12`
= 30 cm2
So, given statement is true, if A coincides Q.
APPEARS IN
RELATED QUESTIONS
In the given figure, ABCD is parallelogram, AE ⊥ DC and CF ⊥ AD. If AB = 16 cm, AE = 8 cm and CF = 10 cm, find AD.
If E, F, G and H are respectively the mid-points of the sides of a parallelogram ABCD show that ar (EFGH) = 1/2ar (ABCD)
In the given below fig. ABCD, ABFE and CDEF are parallelograms. Prove that ar (ΔADE)
= ar (ΔBCF)
In which of the following figures, you find two polygons on the same base and between the same parallels?
In the following figure, PSDA is a parallelogram. Points Q and R are taken on PS such that PQ = QR = RS and PA || QB || RC. Prove that ar (PQE) = ar (CFD).
ABCD is a parallelogram in which BC is produced to E such that CE = BC (Figure). AE intersects CD at F. If ar (DFB) = 3 cm2, find the area of the parallelogram ABCD.
In trapezium ABCD, AB || DC and L is the mid-point of BC. Through L, a line PQ || AD has been drawn which meets AB in P and DC produced in Q (Figure). Prove that ar (ABCD) = ar (APQD)
If the mid-points of the sides of a quadrilateral are joined in order, prove that the area of the parallelogram so formed will be half of the area of the given quadrilateral (Figure).
[Hint: Join BD and draw perpendicular from A on BD.]
ABCD is a trapezium in which AB || DC, DC = 30 cm and AB = 50 cm. If X and Y are, respectively the mid-points of AD and BC, prove that ar (DCYX) = `7/9` ar (XYBA)
In the following figure, ABCD and AEFD are two parallelograms. Prove that ar (PEA) = ar (QFD). [Hint: Join PD].
