Advertisements
Advertisements
प्रश्न
In the following figure, write BC, AC, and CD in ascending order of their lengths.
उत्तर
In ΔABC,
AB = AC
⇒ ∠ABC = ∠ACB ..(angles opposite to equal sides are equal)
⇒ ∠ABC = ∠ACB = 67°
⇒ ∠BAC = 180° - ∠ABC - ∠ACB ...(Angle sum property)
⇒ ∠BAC = 180° - 67° - 67° = 46°
Since ∠BAC < ∠ABC, we have
BC < AC ...(1)
Now, ∠ACD = 180° - ACB ...(Linear pair)
⇒ ∠ACD = 180° - 67° = 113°
Thus, in ΔACD,
∠CAD = 180°- ∠ACD - ∠ADC
⇒ ∠CAD = 180° - 113° - 33° = 34°
Since ∠ADC < ∠CAD, we have
AC < CD ...(2)
From (1) and (2), we have
BC < AC < CD.
APPEARS IN
संबंधित प्रश्न
ABC is a triangle. Locate a point in the interior of ΔABC which is equidistant from all the vertices of ΔABC.
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.
Name the greatest and the smallest sides in the following triangles:
ΔDEF, ∠D = 32°, ∠E = 56° and ∠F = 92°.
Prove that the perimeter of a triangle is greater than the sum of its three medians.
ABCD is a quadrilateral in which the diagonals AC and BD intersect at O. Prove that AB + BC + CD + AD < 2(AC + BC).
In ABC, P, Q and R are points on AB, BC and AC respectively. Prove that AB + BC + AC > PQ + QR + PR.
In the given figure, ∠QPR = 50° and ∠PQR = 60°. Show that: SN < SR
Prove that in an isosceles triangle any of its equal sides is greater than the straight line joining the vertex to any point on the base of the triangle.
In ΔABC, D is a point in the interior of the triangle. Prove that DB + DC < AB + AC.
