vely. PR and SQ are joined. Join AC, PQ, QR, RS and SP.
In ΔABC, P is the mid-point of AB, and Q is the mid-point of BC.
PQ AC and [By mid-point Theorem] …. (1)
In Δ ADC, S is the mid-point of DA and R is the mid-point of DC. [Given] SRAC and [By mid-point Theorem] … (2)
From equations (1) and (2), we get
PQ SR and PQ = SR
Thus, in quadrilateral PQRS, one pair of opposite sides is parallel and equal. PQRS is a parallelogram.
We know that the diagonals of a parallelogram bisect each other so PR and SQ bisect each other.
Hence, the line segments joining the mid-points of the opposite sides of a quadrilateral bisect each other.
30.In the figure given here, CDis parallel to AB and ΔCBA = 25°. Find x.
SOL:- Join AC and BE.
ΔACB = 90° [Angle inscribed in a semi-circle] …. (1)
In ΔACB, we have
∠ACB + ∠CBA + ∠CAB = 180° [Sum of angles of a triangle] ⇒ 90° + 25° + ∠CAB = 180° [Using (1) and ∠CBA = 25°, given] ∠CAB = 180° – 115 °
∠CAB = 65° … (2)
∠CEB = ∠CAB [Angles in the same segment] …. (3)
∠CEB = 65° [Using (2) and (3)] … (4)
CD is parallel AB (given) and CB is the transversa
∠DCB = ∠CBA [Alternate angles] …. (5)
∠CBA = 25° [Given] … (6)
∠DCB = 25° [Using (5) and (6)] …. (7)
∠DEB = ∠DCB [Angles in the same segment] … (8)
∠DEB = 25° [Using (7) and (8) …. (9)
Now x = ∠CED = ∠CEB – ∠DEB
x = 65° – 25° [Using (4) and (9)] x = 40°
Hence, x = 40°.
31. If each diagonal of a quadrilateral separates it into two triangles of equal areas, then show that the quadrilateral is a parallelogram.
Sol:- Let ABCD be the given quadrilateral. Diagonal AC separates it into two equal areas such that ar(ΔACD) = ar(CAB), while diagonal BD separates it into two equal areas such that ar(DAB) = ar(BCD).
Draw CP ⊥ AB produced, and DQ ⊥AB.
ar(quad ABCD) = ar(ΔCAB) + ar(ΔACD)] ar(quad ABCD) = 2ar (ΔCAB) [ ar(ΔACD) = ar(ΔCAB)] ….. (1)
Also, ar (quad ABCD) = ar(ΔDAB) + ar(ΔBCD)
ar(quad ABCD) = 2ar (ΔDAB) [ ar(ΔDAB) = ar(ΔBCD) ….. (2)
From equations (1) and (2), we get
ar(ΔCAB) = ar(ΔDAB)
= 1/ 2 AB X CP = ½ ABX DQ
CP = DQ
Altitude from C of ΔABC = Altitude from D of ΔABD AND DC is parallel AB
Similarly, AD is parallel BC. Hence, ABCD is a parallelogram.
32. The difference between the semi-perimeter and the sides of triangle ABC are 8 cm, 7 cm and 5 cm, respectively. Find its semi-perimeter.
SOL:- Let the semi-perimeter of the triangle be s. Sides of the triangle is a, b and c.
s – a = 8
s – b = 7
s – c = 5
Adding all the equations, we get
(s – a) + (s – b) + (s – c) = 8 + 7 + 5
⇒ 3s – (a + b + c) = 20
⇒ 3s – 2s = 20 AND s = 20 cm
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