1. In the given figure, find the length of QS.
(A) √31
(B) √43
(C) √43
(D) √73
2. For the given figure, which of the following best describes the value of y?
∠POQ=70° and x >15
(A) y<35
(B) y>35
(C) y<55
(D) y>55
3. In given figure, if L∥K, then find the value of y.
(A) 55
(B) 60
(C) 70
(D) 75
4. In the given figure, both triangles are right triangles. Find the area of the shaded region.
(A) 3/5
(B) 7/8
(C) 7/9
(D) 3/7
5. In the given figure, the radius of the larger circle is twice that of the smaller circle. If the circles are concentric, find the ratio of the shaded region’s area to the area of the smaller circle?
(A) 3 : 1
(B) 3 : 5
(C) 7 : 5
(D) 5 : 7
6. In the given figure, ∆PST is an isosceles right triangle, and PS=2. Find the area of the shaded region URST.
(A) 2/3
(B) 5/7
(C) 1/2
(D) 7/9
7. In the given figure, the area of ∆PQR is 40. Find the area of ∆QRS.
(A) 25
(B) 28
(C) 32
(D) 36
8. In the given figure, PQRS is a square and M and N are midpoints of their respective sides. Find the area of the quadrilateral PMRN.
(A) 26
(B) 24
(C) 16
(D) 8
9. In the given figure, what is the greatest number of regions into which two straight lines with divide the shaded region?
(A) 6
(B) 5
(C) 4
(D) 3
10. In the given figure, O is the centre of the circle. If the area of the circle is 9π, then find the perimeter of the sector PRQO.
(A) 1/2 π+6
(B) 1/2 π-6
(C) 2/3 π+6
(D) 2/3 π-6
Answers...
1. C
The length of PR=3+5=8. Applying the Pythagorean theorem to triangle PRS, we get
8^2+(PS)^2=10^2
(PS)^2=100-64=36
∴PS=√36=6
Now, applying the Pythagorean theorem to triangle PQS, we get
(QS)^2=5^2+6^2=25+36=61
∴QS=√61
2.A
Since ∠POQ=70° we get x+y+20=70 Solving this equation for y yields y=50-x. Now, we are given that x>15. Hence, the expression (50-x) must be less than 35. Hence, y<35.
3.D
Since lines L and K are parallel, we know that the corresponding angles are equal. Hence,
y=2y-75
y=75
4.B
Since the height and base of the larger triangle are the same, the slope of the hypotenuse is 45°. Hence, the base of the smaller triangle is the same as its height, 3/2. Thus, the area of the shaded region
=(area of the larger triangle)-(area of the smaller triangle)
=1/2×2×2-1/2×3/2×3/2
=2-9/8=(16-9)/8=7/8
5.A
Let us assume that the radius of the larger circle is 2 and the radius of the smaller circle is 1. Then, the area of the larger circle is πr^2=π(2)^2=4π, and the area of the smaller circle is πr^2=π(1)^2=π. Hence, the area of the shaded region is 4π-π=3π.
∴ (Area of shaded region)/(Area of smaller circle)=3π/π=3/1
6.C
Let, TP=TS=y. Applying the Pythagorean theorem to the right triangle PST, we get
TP^2+TS^2=PS^2
y^2+y^2=2^2
2y^2=4⇒y=√2
Now, area of the shaded region
=area of ∆PST-area of ∆PRU
=1/2×√2×√2-1/2×1×1=1/2
7.A
Area of ∆PQS=1/2×5×6=15
Area of ∆QRS=Area of ∆PQR-Area of ∆PQS
=40-15=25
8.D
Since M is the midpoint of side PQ, the length of MQ is 2. Hence, the are of ∆MQR=1/2×2×4=4. A similar analysis shows that the area of ∆NSR=4. Thus, the unshaded area of the figure=4+4=8
Hence, the area of quadrilateral PMRN
=Area of the square PQRS-The unshaded area of the figure
=16-8=8
9.B
Let us draw the two straight lines as follows.
The lines must intersect in the shaded region.
Hence, the two straight lines will divided the shaded region in 5 parts.
10.A
Since the area of the circle is 9π, we have
πr^2=9π⇒r^2=9⇒r=3
The circumference of the circle is
C=2πr=2π×3=6π
Since the central angle is 30°, the length of arc PRQ
=30/360×C=1/12×6π=1/2 π
Hence, the perimeter of the sector PRQO=1/2 π+3+3=1/2 π+6
1. C
The length of PR=3+5=8. Applying the Pythagorean theorem to triangle PRS, we get
8^2+(PS)^2=10^2
(PS)^2=100-64=36
∴PS=√36=6
Now, applying the Pythagorean theorem to triangle PQS, we get
(QS)^2=5^2+6^2=25+36=61
∴QS=√61
2.A
Since ∠POQ=70° we get x+y+20=70 Solving this equation for y yields y=50-x. Now, we are given that x>15. Hence, the expression (50-x) must be less than 35. Hence, y<35.
3.D
Since lines L and K are parallel, we know that the corresponding angles are equal. Hence,
y=2y-75
y=75
4.B
Since the height and base of the larger triangle are the same, the slope of the hypotenuse is 45°. Hence, the base of the smaller triangle is the same as its height, 3/2. Thus, the area of the shaded region
=(area of the larger triangle)-(area of the smaller triangle)
=1/2×2×2-1/2×3/2×3/2
=2-9/8=(16-9)/8=7/8
5.A
Let us assume that the radius of the larger circle is 2 and the radius of the smaller circle is 1. Then, the area of the larger circle is πr^2=π(2)^2=4π, and the area of the smaller circle is πr^2=π(1)^2=π. Hence, the area of the shaded region is 4π-π=3π.
∴ (Area of shaded region)/(Area of smaller circle)=3π/π=3/1
6.C
Let, TP=TS=y. Applying the Pythagorean theorem to the right triangle PST, we get
TP^2+TS^2=PS^2
y^2+y^2=2^2
2y^2=4⇒y=√2
Now, area of the shaded region
=area of ∆PST-area of ∆PRU
=1/2×√2×√2-1/2×1×1=1/2
7.A
Area of ∆PQS=1/2×5×6=15
Area of ∆QRS=Area of ∆PQR-Area of ∆PQS
=40-15=25
8.D
Since M is the midpoint of side PQ, the length of MQ is 2. Hence, the are of ∆MQR=1/2×2×4=4. A similar analysis shows that the area of ∆NSR=4. Thus, the unshaded area of the figure=4+4=8
Hence, the area of quadrilateral PMRN
=Area of the square PQRS-The unshaded area of the figure
=16-8=8
9.B
Let us draw the two straight lines as follows.
The lines must intersect in the shaded region.
Hence, the two straight lines will divided the shaded region in 5 parts.
10.A
Since the area of the circle is 9π, we have
πr^2=9π⇒r^2=9⇒r=3
The circumference of the circle is
C=2πr=2π×3=6π
Since the central angle is 30°, the length of arc PRQ
=30/360×C=1/12×6π=1/2 π
Hence, the perimeter of the sector PRQO=1/2 π+3+3=1/2 π+6
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