Answer:
1. A. Graph below
1. B. Trapezoid
2. Interior angles are 63.3°, 147.9°, 27.13° and 121.6°.
Explanation:
Ques 1: We are given that, for quadrilateral CONR,
CO is represented by the line
when
![-4\leq x\leq -3](https://img.qammunity.org/2020/formulas/mathematics/middle-school/4f79w85rgyvgg5wdmmydxr69ydj5rofa26.png)
RN is represented by the line
when
![-1\leq x\leq 2](https://img.qammunity.org/2020/formulas/mathematics/middle-school/d96l5pmryma61yqea9c9i6r4cqqcddt8f2.png)
Part A). After plotting the lines, we will get the following graph.
Part B) Joining the end points, we see that, CONR is a trapezoid.
Ques 2: Since, we know,
The sum of the interior angles of a quadrilateral is 360°
So, we have,
![(3x)/(7)+(3x-42)+x+(2x-5)=360](https://img.qammunity.org/2020/formulas/mathematics/middle-school/28qegu0r2avu5xnbin7y5bm3g1385h013e.png)
i.e.
![(3x)/(7)+6x-47=360](https://img.qammunity.org/2020/formulas/mathematics/middle-school/5wf91gyo5ehzif9t09a944z8by2eujhe3y.png)
i.e.
![(3x+42x)/(7)=360+47](https://img.qammunity.org/2020/formulas/mathematics/middle-school/ps4yl4fxs9l3aj9avn9v8208p08q347tur.png)
i.e.
![(45x)/(7)=407](https://img.qammunity.org/2020/formulas/mathematics/middle-school/2qljwpr7se28frqq3pyregx0jc7x35ip1e.png)
i.e.
![x=(407* 7)/(45)](https://img.qammunity.org/2020/formulas/mathematics/middle-school/l11xw54c9u6tzwc45oo5bc5wlv71phrv9m.png)
i.e.
![x=(2849)/(45)](https://img.qammunity.org/2020/formulas/mathematics/middle-school/ksh9mg9yi411gsn6dmbegkokt3qbr66jvf.png)
i.e. x= 63.3°
So, we have,
x= 63.3°
(3x-42)° = (3×63.3 - 42)° = (189.9-42)° = 147.9°
= 27.13°
(2x-5)° = (2×63.3-5)° = (126.6-5)° = 121.6°
Thus, the interior angles are 63.3°, 147.9°, 27.13° and 121.6°.