Advertisements
Advertisements
प्रश्न
In ΔPQR, PS ⊥ QR ; prove that: PQ > QS and PQ > PS
Advertisements
उत्तर
In ΔPQS,
PS < PQ ....(Of all the straight lines that can be drawn to a given straight line from a point outside it, the perpendicular is the shortest.)
I.e. PQ > PS
Also, QS < QP ....(Of all the straight lines that can be drawn to a given straight line from a point outside it, the perpendicular is the shortest.)
i.e. PQ > QS.
APPEARS IN
संबंधित प्रश्न
Complete the hexagonal and star shaped rangolies (see the given figures) by filling them with as many equilateral triangles of side 1 cm as you can. Count the number of triangles in each case. Which has more triangles?
Arrange the sides of ∆BOC in descending order of their lengths. BO and CO are bisectors of angles ABC and ACB respectively.
“Caste inequalities are still prevalent in India.” Examine the statement.
Arrange the sides of the following triangles in an ascending order:
ΔDEF, ∠D = 38°, ∠E = 58°.
D is a point on the side of the BC of ΔABC. Prove that the perimeter of ΔABC is greater than twice of AD.
For any quadrilateral, prove that its perimeter is greater than the sum of its diagonals.
ABCD is a trapezium. Prove that:
CD + DA + AB + BC > 2AC.
In ΔABC, BC produced to D, such that, AC = CD; ∠BAD = 125° and ∠ACD = 105°. Show that BC > CD.
In ΔPQR, PS ⊥ QR ; prove that: PQ + PR > QR and PQ + QR >2PS.
In the given figure, T is a point on the side PR of an equilateral triangle PQR. Show that PT < QT
