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Practice each question, then open the answer or worked solution.
The median of a triangle divides it into two
(A) triangles of equal area
(B) congruent triangles
(C) right triangles
(D) isosceles triangles
(A) triangles of equal area
Explanation:
The median of a triangle divides it into triangles of equal area.
Hence, option (A) is the correct answer.
In which of the following figures (Fig. 9.3) do you find two polygons on the same base and between the same parallels?
(D)
Explanation:
In figure (D), the parallelograms, PQRA and BQRS, are on the same base QR and between the same parallels QR and PS.
Hence, option (D) is the correct answer.
The figure obtained by joining the mid-points of the adjacent sides of a rectangle of sides 8 cm and 6 cm is :
(A) a rectangle of area 24 cm2
(B) a square of area 25 cm2
(C) a trapezium of area 24 cm2
(D) a rhombus of area 24 cm2
(D) a rhombus of area 24 cm2
Explanation:
According to the question,
Let ABCD be the rectangle.
E is the mid-point of the side AB
F is the mid-point of the side BC
G is the mid-point of the side CD
H is the mid-point of the side DA
The figure obtained by joining the midpoints E, F, G and H is a rhombus.
The sides of the rectangle are given,
We know that,
Side of the rectangle AB = diagonal of the rhombus FH = 8cm
Similarly,
Side of the rectangle AD = diagonal of the rhombus EG = 6cm
Then,
Area of the rhombus = ½ × EG × FH
= ½ × 6 × 8
= 24 cm2
Hence, option (D) is the correct answer.
In Fig. 9.4, the area of parallelogram
ABCD is:
(A) AB × BM
(B) BC × BN
(C) DC × DL
(D) AD × DL
(C) DC × DL
Explanation:
Area of parallelogram = Base × Corresponding altitude
= AB × DL … (eq 1)
Since opposite sides of a parallelogram are equal,
We get,
AB = DC
Substituting this in eq(1), we get,
Area of parallelogram = AB × DL
= DC × DL
Hence, option (C) is the correct answer.
In Fig. 9.5, if parallelogram ABCD and rectangle ABEF are of equal area, then :
(A) Perimeter of ABCD = Perimeter of ABEM
(B) Perimeter of ABCD < Perimeter of ABEM
(C) Perimeter of ABCD > Perimeter of ABEM
(D) Perimeter of ABCD = ½ (Perimeter of ABEM)
(C) Perimeter of ABCD > Perimeter of ABEM
Explanation:
In rectangle ABEM,
AB = EM …(eq.1) [sides of rectangle]
In parallelogram ABCD,
CD = AB …(eq.2)
Adding equations (1) and (2),
We get
AB + CD = EM + AB …(i)
We know that,
The perpendicular distance between two parallel sides of a parallelogram is always less than the length of the other parallel sides.
BE < BC and AM < AD
On adding both above inequalities, we get
SE + AM <BC + AD or BC + AD> BE + AM
On adding AB + CD on both sides, we get
AB + CD + BC + AD> AB + CD + BE + AM
⇒ AB+BC + CD + AD> AB+BE + EM+ AM [∴ CD = AB = EM]
Hence,
We get,
The perimeter of parallelogram ABCD > perimeter of rectangle ABEM
Hence, option (C) is the correct answer.