Advertisements
Advertisements
Question
The figure, given below, shows a trapezium ABCD. M and N are the mid-point of the non-parallel sides AD and BC respectively. Find:
- MN, if AB = 11 cm and DC = 8 cm.
- AB, if DC = 20 cm and MN = 27 cm.
- DC, if MN = 15 cm and AB = 23 cm.
Advertisements
Solution
Let we draw a diagonal AC as shown in the figure below,
(i) Given that AB = 11 cm, CD = 8 cm
From triangle ABC
ON = `[1]/[2]` AB
= `[1]/[2]` × 11
= 5.5 cm
From triangle ACD
OM = `[1]/[2]` CD
=`[1]/[2]` × 8
= 4 cm
Hence, MN = OM + ON
= (4 + 5.5)
= 9.5 cm
(ii) Given that CD = 20 cm, MN = 27 cm
From triangle ACD
OM = `[1]/[2]` CD
= `[1]/[2]` × 20
= 10 cm
Therefore, ON = 27 - 10 = 17 cm
From triangle ABC
AB = 2ON
= 2 × 17
= 34 cm
(iii) Given that AB = 23 cm, MN = 15 cm
From triangle ABC
ON =`[1]/[2]` AB
=`[1]/[2]` × 23
= 11.5 cm
OM = 15 - 11.5
OM = 3.5 cm
From triangle ACD
CD = 2OM
= 2 × 3.5
CD = 7 cm
APPEARS IN
RELATED QUESTIONS
ABCD is a rhombus and P, Q, R and S are the mid-points of the sides AB, BC, CD and DA respectively. Show that the quadrilateral PQRS is a rectangle.
ABCD is a rhombus. EABF is a straight line such that EA = AB = BF. Prove that ED and FC when produced, meet at right angles.
In a triangle ∠ABC, ∠A = 50°, ∠B = 60° and ∠C = 70°. Find the measures of the angles of
the triangle formed by joining the mid-points of the sides of this triangle.
In a ΔABC, E and F are the mid-points of AC and AB respectively. The altitude AP to BC
intersects FE at Q. Prove that AQ = QP.
In the Figure, `square`ABCD is a trapezium. AB || DC. Points P and Q are midpoints of seg AD and seg BC respectively. Then prove that, PQ || AB and PQ = `1/2 ("AB" + "DC")`.
In ∆ABC, E is the mid-point of the median AD, and BE produced meets side AC at point Q.
Show that BE: EQ = 3: 1.
If the quadrilateral formed by joining the mid-points of the adjacent sides of quadrilateral ABCD is a rectangle,
show that the diagonals AC and BD intersect at the right angle.
In ΔABC, BE and CF are medians. P is a point on BE produced such that BE = EP and Q is a point on CF produced such that CF = FQ. Prove that: QAP is a straight line.
ABCD is a kite in which BC = CD, AB = AD. E, F and G are the mid-points of CD, BC and AB respectively. Prove that: ∠EFG = 90°
In the given figure, T is the midpoint of QR. Side PR of ΔPQR is extended to S such that R divides PS in the ratio 2:1. TV and WR are drawn parallel to PQ. Prove that T divides SU in the ratio 2:1 and WR = `(1)/(4)"PQ"`.
