D, E, F are midpoints of sides AB, BC and CA of ΔABC, if ar(ΔABC) = 64 cm2 then, area of ΔBDE is:
Bcs (ABC) is a full triangle ABC triangle divide into four equal partsTherefore, 64÷4=16 cm2
In the adjoining figure, ABCD and PQRC are rectangles, where Q is the midpoint of AC. Then DP is equal to
Given that PQRS and ABCD are rectangle and Q is the mid pt of AC.
Since PQRS and ABCD are rectangle, every angle would be of 90°.
Consider AD and PQ.
angle ADP = 90°
angle QPC = 90°
Now, angle ADP lies corresponding to angle QPC and are equal.
If corresponding angles are equal, then the lines are parallel
=> AD || PQ
Now consider ∆ADC.
Q is the midpoint of AC (Given)
AD || PQ (shown above)
We know that,
line drawn parallel from the midpoint of a side, to other side, bisects the third side.
Hence, P would be the midpoint of DC
Hence, DP = PC
In triangle ABC, E and F are the mid points of the sides AC and AB respectively. The altitude AP to BC intersects EF at Q. Then
Figure shows that AD and BF are medians of ABC and BF || DE then CE is equal to
A median divides the triangle into 2 equal parts. So AD divides the triangle into the triangle ABD & ACD . Further BF divides the triangle into 2 equal parts .
So AF = 1/2 AC
DE bisects it into 2 parts
So EF = 1/2 EC
AF = EC
SO FC =1/2 * 1/2 AC
= 1/4 AC
In ΔABC, D, E and F are respectively mid points of BC, CA and AB. If the lengths of side AB, BC and CA are 7 cm, 8 cm and 9 cm respectively, the perimeter of DEF will be
Since, D,E,F are the midpoints of BC,CA and AB,Thus
If D, E and F are the mid points of the sides BC, CA and AB of an equilateral triangle ABC, then triangle DEF is
The quadrilateral formed by joining the midpoints of the pairs of adjacent sides of a rectangle is a
ABCD is a parallelogram in which P, Q, R and S are the mid points of sides AB, BC CD and DA respectively. Then,
Find the area of a trapezium whose parallel sides are 24 cm and 20 cm and the distance between them is 15 cm.
In the adjoining figure, the side AC of ABC is produced to E such that CE = 1/2 AC. If D is the midpoint of BC and ED produced meets AB at F, and CP, DQ are drawn parallel to BA, then FD is
ABC is a triangle.
D is midpoint of BC and DQ is drawn parallel to BA.
Then, Q is midpoint of AC.
∴ AQ = DC
∴ FA parallel to DQ||PC.
AQC, is a transversal so, AQ = QC and FDP also a transveral on them.
∴ FD = DP .......(1) [ intercept theorem]
EC = 1/2 AC = QC
Now, triangle EQD, here C is midpoint of EQ and CP which is parallel to DQ.
And, P is midpoint of DE.
DP = PE..........(2)
Therefore, From (1) and (2)
FD = DP = PE
∴ FD = 1/3 FE