Advertisements
Advertisements
Question
In ∆ABC, AB = 5 cm, BC = 8 cm and CA = 7 cm. If D and E are respectively the mid-points of AB and BC, determine the length of DE.
Advertisements
Solution
In ΔABC, we have AB = 5 cm, BC = 8 cm and CA = 7 cm.
Since, D and E are the mid-points of AB and BC, respectively.
By mid-point theorem, DE || AC
And DE = `1/2` AC = `7/2` = 3.5 cm

APPEARS IN
RELATED QUESTIONS
Show that the line segments joining the mid-points of the opposite sides of a quadrilateral bisect each other.
In triangle ABC, the medians BP and CQ are produced up to points M and N respectively such that BP = PM and CQ = QN. Prove that:
- M, A, and N are collinear.
- A is the mid-point of MN.
In ΔABC, AB = 12 cm and AC = 9 cm. If M is the mid-point of AB and a straight line through M parallel to AC cuts BC in N, what is the length of MN?
In ΔABC, D, E, F are the midpoints of BC, CA and AB respectively. Find ∠FDB if ∠ACB = 115°.
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: A is the mid-point of PQ.
In parallelogram ABCD, P is the mid-point of DC. Q is a point on AC such that CQ = `(1)/(4)"AC"`. PQ produced meets BC at R. Prove that
(i) R is the mid-point of BC, and
(ii) PR = `(1)/(2)"DB"`.
In the given figure, ABCD is a trapezium. P and Q are the midpoints of non-parallel side AD and BC respectively. Find: DC, if AB = 20 cm and PQ = 14 cm
In a parallelogram ABCD, E and F are the midpoints of the sides AB and CD respectively. The line segments AF and BF meet the line segments DE and CE at points G and H respectively Prove that: EGFH is a parallelogram.
D, E and F are the mid-points of the sides BC, CA and AB, respectively of an equilateral triangle ABC. Show that ∆DEF is also an equilateral triangle.
E and F are respectively the mid-points of the non-parallel sides AD and BC of a trapezium ABCD. Prove that EF || AB and EF = `1/2` (AB + CD).
[Hint: Join BE and produce it to meet CD produced at G.]
