Answer:
1. Answer: (b) Base × Altitude
2. Answer: (c) 1 :1
3. Answer: (c) 12cm
Explanation: Given,
AB = CD = 10 cm (Opposite sides of a parallelogram)
CF = 5 cm and AE = 6 cm
Now,
Area of parallelogram = Base × Altitude
CD × AE = AD × CF
10 × 6 = AD × 5
AD = 60/5
AD = 12 cm
4. Answer: (c) ar(APB) = ar(BQC)
Explanation: ΔAPB and parallelogram ABCD lie on the same base AB and in-between same parallel AB and DC.
ar(ΔAPB) = 1/2 ar(parallelogram ABCD) — 1
ar(ΔBQC) = 1/2 ar(parallelogram ABCD) — 2
From eq. 1 and 2:
ar(ΔAPB) = ar(ΔBQC)
5. Answer: (c) ar (ABCD) = ar (EFGH)
6. Answer: (d) Equal area triangles
Explanation: Suppose, ABC is a triangle and AD is the median.
AD is the median of ΔABC.
∴ It will divide ΔABC into two triangles of equal area.
∴ ar(ABD) = ar(ACD) — (i)
also,
ED is the median of ΔABC.
∴ ar(EBD) = ar(ECD) — (ii)
Subtracting (ii) from (i),
ar(ABD) – ar(EBD) = ar(ACD) – ar(ECD)
⇒ ar(ABE) = ar(ACE)
7. Answer: (b) DE is parallel to BC
Explanation: ΔDBC and ΔEBC are on the same base BC and also having equal areas.
∴ They will lie between the same parallel lines.
∴ DE || BC.
8. Answer: (a) ar (AOD) = ar (BOC)
Explanation: △DAC and △DBC lie on the same base DC and between the same parallels AB and CD.
ar(△DAC) = ar(△DBC)
⇒ ar(△DAC) − ar(△DOC) = ar(△DBC) − ar(△DOC)
⇒ ar(△AOD) = ar(△BOC)
9. Answer: (d) Trapezium
Explanation: ar(△AOD) = ar(△BOC)
ar(△AOD) = ar(△BOC)
⇒ ar(△AOD) + ar(△AOB) = ar(△BOC) + ar(△AOB)
⇒ ar(△ADB) = ar(△ACB)
Areas of △ADB and △ACB are equal.
Therefore, they must lie between the same parallel lines.
Therefore, AB ∥ CD
Hence, ABCD is a trapezium.
10. Answer: (b) The line containing the common base
11. Answer: (b) 1 : 2
Explanation: The ratio of area of a triangle and rhombus is 1:2. Since rhombus is a parallelogram and the triangle and rhombus lie on same base and between same parallel.
∴ar(triangle)= 1/2 ar(rhombus)
∴ The ratio =1:2
12. Answer: (b) (1/2) ar(ABC)
Explanation: Now, consider a triangle ABC as shown in the figure.
Let the area of the triangle be “A”.
By using Heron’s formula
Therefore, the area of triangle AEF = A/4
Area of triangle CFG = A/4
Therefore, the area of parallelogram BEFG = Area of triangle ABC – Area of AEF – Area of CFG
= A – (A/4) – (A/4)
= (4A – A – A)/4
= 2A/4
=(1/2) A.
13. Answer: (c) DC × DL
Explanation: The area of a parallelogram for the given figure is DC × DL, as the formula for the area of the parallelogram is base×height.
14. Answer: (d) triangles of equal areas
15. Answer: (c) 13 cm
Explanation: We know that, area of triangle = ½ × base (small side) × height
30 = 1/2 × 5 cm × height
Height = (30 × 2)/5
Height = 60/5
Height = 12 cm
By the rule of Pythagoras theorem,
Hypotenuse2 = height2 + base2
Hypotenuse2 = 122 + 52
Hypotenuse2 = 144 + 25
Hypotenuse2 = 169
Hypotenuse = √169
Hypotenuse = 13 cm
16. Answer: (a) 1:2
17. Answer: (b) Quadrilateral
Explanation: Parallelogram is a quadrilateral, as it has four sides.
18. Answer: (a) 1: 1
Explanation: We know that the parallelogram on the equal bases and between the same parallels are equal in area. Hence, the ratio of their areas is 1 :1.
19. Answer: (a) b×h square units
Explanation: The area of a parallelogram with base “b” and height “h” is b×h.(i.e) A = base × height square units.
20. Answer: (b) Equal in area
Explanation: Triangles on the same base and between the same parallels are equal in area.
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